{"id":458,"date":"2025-07-22T08:38:03","date_gmt":"2025-07-22T08:38:03","guid":{"rendered":"https:\/\/giapsgk.com\/chuyen-de\/?p=458"},"modified":"2025-07-22T09:57:48","modified_gmt":"2025-07-22T09:57:48","slug":"cac-cong-thuc-toan-lop-10","status":"publish","type":"post","link":"https:\/\/giapsgk.com\/chuyen-de\/cac-cong-thuc-toan-lop-10\/","title":{"rendered":"H\u1ec7 th\u1ed1ng c\u00e1c c\u00f4ng th\u1ee9c To\u00e1n 10 (h\u1ecdc k\u00ec 1 v\u00e0 h\u1ecdc k\u00ec 2 &#8211; s\u00e1ch m\u1edbi)"},"content":{"rendered":"<p data-start=\"482\" data-end=\"795\">To\u00e1n h\u1ecdc l\u1edbp 10 l\u00e0 b\u01b0\u1edbc kh\u1edfi \u0111\u1ea7u quan tr\u1ecdng trong ch\u01b0\u01a1ng tr\u00ecnh trung h\u1ecdc ph\u1ed5 th\u00f4ng, n\u01a1i h\u1ecdc sinh l\u00e0m quen v\u1edbi nh\u1eefng kh\u00e1i ni\u1ec7m tr\u1eebu t\u01b0\u1ee3ng h\u01a1n v\u00e0 t\u01b0 duy logic s\u00e2u s\u1eafc h\u01a1n. Vi\u1ec7c n\u1eafm ch\u1eafc <strong data-start=\"666\" data-end=\"699\">c\u00e1c c\u00f4ng th\u1ee9c to\u00e1n h\u1ecdc l\u1edbp 10<\/strong> kh\u00f4ng ch\u1ec9 gi\u00fap c\u00e1c em h\u1ecdc t\u1ed1t hi\u1ec7n t\u1ea1i, m\u00e0 c\u00f2n t\u1ea1o n\u1ec1n t\u1ea3ng v\u1eefng ch\u1eafc cho l\u1edbp 11 v\u00e0 12 sau n\u00e0y.<\/p>\n<p data-start=\"797\" data-end=\"1182\">V\u1edbi vai tr\u00f2 l\u00e0 m\u1ed9t nh\u00e0 gi\u00e1o d\u1ee5c, t\u00f4i th\u1ea5u hi\u1ec3u r\u1eb1ng nhi\u1ec1u h\u1ecdc sinh c\u1ea3m th\u1ea5y kh\u00f3 kh\u0103n khi nh\u1edb h\u1ebft c\u00e1c c\u00f4ng th\u1ee9c r\u1ea3i r\u00e1c trong s\u00e1ch gi\u00e1o khoa. Ch\u00ednh v\u00ec v\u1eady, b\u00e0i vi\u1ebft n\u00e0y \u0111\u01b0\u1ee3c bi\u00ean so\u1ea1n nh\u1eb1m <strong data-start=\"985\" data-end=\"1025\">h\u1ec7 th\u1ed1ng \u0111\u1ea7y \u0111\u1ee7, khoa h\u1ecdc v\u00e0 d\u1ec5 hi\u1ec3u<\/strong> t\u1ea5t c\u1ea3 c\u00e1c c\u00f4ng th\u1ee9c quan tr\u1ecdng c\u1ee7a To\u00e1n l\u1edbp 10 \u2013 t\u1eeb <strong data-start=\"1079\" data-end=\"1100\">M\u1ec7nh \u0111\u1ec1 \u2013 T\u1eadp h\u1ee3p<\/strong>, <strong data-start=\"1102\" data-end=\"1131\">H\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t \u2013 b\u1eadc hai<\/strong>, \u0111\u1ebfn <strong data-start=\"1137\" data-end=\"1146\">Vect\u01a1<\/strong>, <strong data-start=\"1148\" data-end=\"1174\">T\u1ecda \u0111\u1ed9 trong m\u1eb7t ph\u1eb3ng<\/strong>, v.v.<\/p>\n<p data-start=\"1184\" data-end=\"1321\">T\u00e0i li\u1ec7u ph\u00f9 h\u1ee3p cho c\u1ea3 h\u1ecdc sinh mu\u1ed1n \u00f4n t\u1eadp, ghi nh\u1edb ki\u1ebfn th\u1ee9c tr\u1ecdng t\u00e2m, v\u00e0 gi\u00e1o vi\u00ean c\u1ea7n t\u1ed5ng h\u1ee3p \u0111\u1ec3 gi\u1ea3ng d\u1ea1y ho\u1eb7c ki\u1ec3m tra \u0111\u00e1nh gi\u00e1.<\/p>\n<h2><strong>Ch\u01b0\u01a1ng I: M\u1ec7nh \u0111\u1ec1 v\u00e0 T\u1eadp h\u1ee3p<\/strong><\/h2>\n<h3><strong>1. M\u1ec7nh \u0111\u1ec1<\/strong><\/h3>\n<ul>\n<li><strong>\u0110\u1ecbnh ngh\u0129a<\/strong>: M\u1ec7nh \u0111\u1ec1 l\u00e0 m\u1ed9t c\u00e2u kh\u1eb3ng \u0111\u1ecbnh c\u00f3 gi\u00e1 tr\u1ecb \u0111\u00fang ho\u1eb7c sai, kh\u00f4ng \u0111\u1ed3ng th\u1eddi \u0111\u00fang v\u00e0 sai. V\u00ed d\u1ee5: &#8220;4 l\u00e0 s\u1ed1 ch\u1eb5n&#8221; (\u0111\u00fang), &#8220;3 &gt; 5&#8221; (sai).<\/li>\n<li><strong>Ghi ch\u00fa quan tr\u1ecdng<\/strong>:\n<ul>\n<li>Ph\u00e9p ph\u1ee7 \u0111\u1ecbnh c\u1ee7a m\u1ec7nh \u0111\u1ec1 $p$ l\u00e0 $\\neg p$, c\u00f3 gi\u00e1 tr\u1ecb ng\u01b0\u1ee3c l\u1ea1i.<\/li>\n<li>M\u1ec7nh \u0111\u1ec1 k\u00e9o theo $p \\to q$: N\u1ebfu $p$ \u0111\u00fang th\u00ec $q$ ph\u1ea3i \u0111\u00fang.<\/li>\n<li>M\u1ec7nh \u0111\u1ec1 t\u01b0\u01a1ng \u0111\u01b0\u01a1ng $p \\leftrightarrow q$: $p$ v\u00e0 $q$ c\u00f3 c\u00f9ng gi\u00e1 tr\u1ecb \u0111\u00fang\/sai.<\/li>\n<\/ul>\n<\/li>\n<li><strong>C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\n<ul>\n<li>Ph\u00e9p h\u1ed9i: $p \\land q$ (\u0111\u00fang khi c\u1ea3 $p$ v\u00e0 $q$ \u0111\u00fang).<\/li>\n<li>Ph\u00e9p tuy\u1ec3n: $p \\lor q$ (\u0111\u00fang khi \u00edt nh\u1ea5t m\u1ed9t trong $p$ ho\u1eb7c $q$ \u0111\u00fang).<\/li>\n<li>Ph\u00e9p k\u00e9o theo: $p \\to q \\equiv \\neg p \\lor q$.<\/li>\n<li>Ph\u00e9p t\u01b0\u01a1ng \u0111\u01b0\u01a1ng: $p \\leftrightarrow q \\equiv (p \\to q) \\land (q \\to p)$.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<h3><strong>2. T\u1eadp h\u1ee3p v\u00e0 c\u00e1c ph\u00e9p to\u00e1n tr\u00ean t\u1eadp h\u1ee3p<\/strong><\/h3>\n<ul>\n<li><strong>\u0110\u1ecbnh ngh\u0129a<\/strong>:\n<ul>\n<li>T\u1eadp h\u1ee3p l\u00e0 t\u1eadp h\u1ee3p c\u00e1c ph\u1ea7n t\u1eed c\u00f3 t\u00ednh ch\u1ea5t chung. K\u00fd hi\u1ec7u: $A, B, \\ldots$.<\/li>\n<li>T\u1eadp h\u1ee3p con: $A \\subseteq B$ n\u1ebfu m\u1ecdi ph\u1ea7n t\u1eed c\u1ee7a $A$ thu\u1ed9c $B$.<\/li>\n<li>T\u1eadp h\u1ee3p l\u0169y th\u1eeba: T\u1eadp h\u1ee3p t\u1ea5t c\u1ea3 t\u1eadp con c\u1ee7a t\u1eadp h\u1ee3p $A$.<\/li>\n<\/ul>\n<\/li>\n<li><strong>Ghi ch\u00fa quan tr\u1ecdng<\/strong>:\n<ul>\n<li>T\u1eadp r\u1ed7ng: $\\emptyset$, kh\u00f4ng ch\u1ee9a ph\u1ea7n t\u1eed n\u00e0o.<\/li>\n<li>S\u1ed1 ph\u1ea7n t\u1eed c\u1ee7a t\u1eadp l\u0169y th\u1eeba: N\u1ebfu $|A| = n$, th\u00ec $|\\mathcal{P}(A)| = 2^n$.<\/li>\n<\/ul>\n<\/li>\n<li><strong>C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\n<ul>\n<li>Ph\u00e9p h\u1ee3p: $A \\cup B = { x \\mid x \\in A \\text{ ho\u1eb7c } x \\in B }$.<\/li>\n<li>Ph\u00e9p giao: $A \\cap B = { x \\mid x \\in A \\text{ v\u00e0 } x \\in B }$.<\/li>\n<li>Ph\u00e9p hi\u1ec7u: $A \\setminus B = { x \\mid x \\in A \\text{ v\u00e0 } x \\notin B }$.<\/li>\n<li>Ph\u00e9p b\u00f9: $\\overline{A} = { x \\mid x \\in U \\text{ v\u00e0 } x \\notin A }$, v\u1edbi $U$ l\u00e0 t\u1eadp h\u1ee3p t\u1ed5ng qu\u00e1t.<\/li>\n<li>C\u00f4ng th\u1ee9c \u0111\u1ebfm: $|A \\cup B| = |A| + |B| &#8211; |A \\cap B|$.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-468 alignnone size-medium aligncenter\" src=\"https:\/\/giapsgk.com\/chuyen-de\/wp-content\/uploads\/giapsgk-cac-cong-thuc-toan-lop-10-2.png\" alt=\"H\u00ecnh \u1ea3nh - cac-cong-thuc-toan-lop-10\" width=\"501\" height=\"284\" \/><\/p>\n<h2><strong>Ch\u01b0\u01a1ng II: B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh v\u00e0 H\u1ec7 b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t hai \u1ea9n<\/strong><\/h2>\n<h3><strong>1. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t hai \u1ea9n<\/strong><\/h3>\n<ul>\n<li><strong>\u0110\u1ecbnh ngh\u0129a<\/strong>: L\u00e0 b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh d\u1ea1ng $ax + by + c &gt; 0$ (ho\u1eb7c $\\geq, &lt;, \\leq$).<\/li>\n<li><strong>Ghi ch\u00fa quan tr\u1ecdng<\/strong>:\n<ul>\n<li>Mi\u1ec1n nghi\u1ec7m l\u00e0 n\u1eeda m\u1eb7t ph\u1eb3ng \u0111\u01b0\u1ee3c x\u00e1c \u0111\u1ecbnh b\u1edfi \u0111\u01b0\u1eddng th\u1eb3ng $ax + by + c = 0$.<\/li>\n<li>Ki\u1ec3m tra m\u1ed9t \u0111i\u1ec3m \u0111\u1ec3 x\u00e1c \u0111\u1ecbnh n\u1eeda m\u1eb7t ph\u1eb3ng nghi\u1ec7m.<\/li>\n<\/ul>\n<\/li>\n<li><strong>C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\n<ul>\n<li>Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng: $ax + by + c = 0$.<\/li>\n<li>Mi\u1ec1n nghi\u1ec7m: Thay m\u1ed9t \u0111i\u1ec3m v\u00e0o b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh \u0111\u1ec3 x\u00e1c \u0111\u1ecbnh v\u00f9ng.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<h3><strong>2. H\u1ec7 b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t hai \u1ea9n<\/strong><\/h3>\n<ul>\n<li><strong>\u0110\u1ecbnh ngh\u0129a<\/strong>: H\u1ec7 g\u1ed3m nhi\u1ec1u b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t hai \u1ea9n, mi\u1ec1n nghi\u1ec7m l\u00e0 giao c\u00e1c n\u1eeda m\u1eb7t ph\u1eb3ng nghi\u1ec7m.<\/li>\n<li><strong>Ghi ch\u00fa quan tr\u1ecdng<\/strong>:\n<ul>\n<li>Mi\u1ec1n nghi\u1ec7m th\u01b0\u1eddng l\u00e0 \u0111a gi\u00e1c l\u1ed3i.<\/li>\n<li>B\u00e0i to\u00e1n t\u1ed1i \u01b0u: T\u00ecm gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t\/nh\u1ecf nh\u1ea5t c\u1ee7a h\u00e0m m\u1ee5c ti\u00eau $f(x, y) = ax + by$.<\/li>\n<\/ul>\n<\/li>\n<li><strong>C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\n<ul>\n<li>Mi\u1ec1n nghi\u1ec7m: Giao c\u00e1c n\u1eeda m\u1eb7t ph\u1eb3ng ${ (x, y) \\mid ax_i + by_i + c_i \\geq 0, i = 1, 2, \\ldots }$.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<h2><strong>Ch\u01b0\u01a1ng III: H\u00e0m s\u1ed1 v\u00e0 \u0110\u1ed3 th\u1ecb<\/strong><\/h2>\n<h3><strong>1. H\u00e0m s\u1ed1 v\u00e0 \u0110\u1ed3 th\u1ecb<\/strong><\/h3>\n<ul>\n<li><strong>\u0110\u1ecbnh ngh\u0129a<\/strong>: H\u00e0m s\u1ed1 $y = f(x)$ l\u00e0 \u00e1nh x\u1ea1 t\u1eeb t\u1eadp x\u00e1c \u0111\u1ecbnh $D$ sang t\u1eadp gi\u00e1 tr\u1ecb.<\/li>\n<li><strong>Ghi ch\u00fa quan tr\u1ecdng<\/strong>:\n<ul>\n<li>T\u00ednh ch\u1ea5t: \u0110\u1ed3ng bi\u1ebfn ($x_1 &lt; x_2 \\implies f(x_1) &lt; f(x_2)$), ngh\u1ecbch bi\u1ebfn, ch\u1eb5n, l\u1ebb, tu\u1ea7n ho\u00e0n.<\/li>\n<li>\u0110\u1ed3 th\u1ecb: T\u1eadp h\u1ee3p c\u00e1c \u0111i\u1ec3m $(x, f(x))$ tr\u00ean m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9.<\/li>\n<\/ul>\n<\/li>\n<li><strong>C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\n<ul>\n<li>T\u1eadp x\u00e1c \u0111\u1ecbnh: T\u1eadp $D$ sao cho $f(x)$ c\u00f3 ngh\u0129a.<\/li>\n<li>\u0110\u1ed3ng bi\u1ebfn: $f'(x) &gt; 0$, ngh\u1ecbch bi\u1ebfn: $f'(x) &lt; 0$ (n\u1ebfu h\u00e0m kh\u1ea3 vi).<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<h3><strong>2. H\u00e0m s\u1ed1 b\u1eadc hai<\/strong><\/h3>\n<ul>\n<li><strong>\u0110\u1ecbnh ngh\u0129a<\/strong>: H\u00e0m s\u1ed1 d\u1ea1ng $y = ax^2 + bx + c$ ($a \\neq 0$).<\/li>\n<li><strong>Ghi ch\u00fa quan tr\u1ecdng<\/strong>:\n<ul>\n<li>\u0110\u1ed3 th\u1ecb l\u00e0 parabol, m\u1edf l\u00ean n\u1ebfu $a &gt; 0$, m\u1edf xu\u1ed1ng n\u1ebfu $a &lt; 0$.<\/li>\n<li>\u0110\u1ec9nh parabol: $x = -\\frac{b}{2a}$, $y = f\\left(-\\frac{b}{2a}\\right)$.<\/li>\n<\/ul>\n<\/li>\n<li><strong>C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\n<ul>\n<li>\u0110\u1ec9nh: $\\left(-\\frac{b}{2a}, \\frac{4ac &#8211; b^2}{4a}\\right)$.<\/li>\n<li>Tr\u1ee5c \u0111\u1ed1i x\u1ee9ng: $x = -\\frac{b}{2a}$.<\/li>\n<li>Gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t\/nh\u1ecf nh\u1ea5t: T\u1ea1i \u0111\u1ec9nh parabol.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<h3><strong>3. D\u1ea5u c\u1ee7a tam th\u1ee9c b\u1eadc hai<\/strong><\/h3>\n<ul>\n<li><strong>\u0110\u1ecbnh ngh\u0129a<\/strong>: X\u00e9t d\u1ea5u c\u1ee7a $f(x) = ax^2 + bx + c$.<\/li>\n<li><strong>Ghi ch\u00fa quan tr\u1ecdng<\/strong>:\n<ul>\n<li>D\u1ef1a v\u00e0o nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $ax^2 + bx + c = 0$ v\u00e0 d\u1ea5u c\u1ee7a $a$.<\/li>\n<li>S\u1eed d\u1ee5ng b\u1ea3ng x\u00e9t d\u1ea5u \u0111\u1ec3 x\u00e1c \u0111\u1ecbnh kho\u1ea3ng nghi\u1ec7m.<\/li>\n<\/ul>\n<\/li>\n<li><strong>C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\n<ul>\n<li>Nghi\u1ec7m: $x_{1,2} = \\frac{-b \\pm \\sqrt{b^2 &#8211; 4ac}}{2a}$, v\u1edbi $\\Delta = b^2 &#8211; 4ac$.<\/li>\n<li>D\u1ea5u c\u1ee7a tam th\u1ee9c:\n<ul>\n<li>$\\Delta &lt; 0$: D\u1ea5u kh\u00f4ng \u0111\u1ed5i, c\u00f9ng d\u1ea5u v\u1edbi $a$.<\/li>\n<li>$\\Delta \\geq 0$: X\u00e9t d\u1ea5u theo nghi\u1ec7m $x_1, x_2$.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<h3><strong>4. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai m\u1ed9t \u1ea9n<\/strong><\/h3>\n<ul>\n<li><strong>\u0110\u1ecbnh ngh\u0129a<\/strong>: B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh d\u1ea1ng $ax^2 + bx + c &gt; 0$ (ho\u1eb7c $\\geq, &lt;, \\leq$).<\/li>\n<li><strong>Ghi ch\u00fa quan tr\u1ecdng<\/strong>:\n<ul>\n<li>K\u1ebft h\u1ee3p nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh v\u00e0 b\u1ea3ng x\u00e9t d\u1ea5u.<\/li>\n<\/ul>\n<\/li>\n<li><strong>C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\n<ul>\n<li>Nghi\u1ec7m c\u1ee7a $ax^2 + bx + c = 0$: $x_{1,2} = \\frac{-b \\pm \\sqrt{b^2 &#8211; 4ac}}{2a}$.<\/li>\n<li>B\u1ea3ng x\u00e9t d\u1ea5u \u0111\u1ec3 t\u00ecm kho\u1ea3ng nghi\u1ec7m.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<h3><strong>5. Ph\u01b0\u01a1ng tr\u00ecnh quy v\u1ec1 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai<\/strong><\/h3>\n<ul>\n<li><strong>\u0110\u1ecbnh ngh\u0129a<\/strong>: Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 th\u1ec3 bi\u1ebfn \u0111\u1ed5i v\u1ec1 d\u1ea1ng $ax^2 + bx + c = 0$ b\u1eb1ng c\u00e1ch \u0111\u1eb7t \u1ea9n ph\u1ee5.<\/li>\n<li><strong>Ghi ch\u00fa quan tr\u1ecdng<\/strong>:\n<ul>\n<li>C\u00e1c d\u1ea1ng ph\u1ed5 bi\u1ebfn: Ph\u01b0\u01a1ng tr\u00ecnh l\u01b0\u1ee1ng th\u1ee9c, ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc b\u1ed1n, ph\u01b0\u01a1ng tr\u00ecnh ch\u1ee9a c\u0103n.<\/li>\n<\/ul>\n<\/li>\n<li><strong>C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\n<ul>\n<li>\u0110\u1eb7t $t = g(x)$, gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai theo $t$, sau \u0111\u00f3 t\u00ecm $x$.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<h2><strong>Ch\u01b0\u01a1ng IV: H\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong tam gi\u00e1c<\/strong><\/h2>\n<h3><strong>1. Gi\u00e1 tr\u1ecb l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a m\u1ed9t g\u00f3c t\u1eeb 0$^0$ \u0111\u1ebfn 180<\/strong><strong>$^0$<\/strong><\/h3>\n<ul>\n<li><strong>\u0110\u1ecbnh ngh\u0129a<\/strong>: C\u00e1c h\u00e0m s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c $\\sin \\alpha, \\cos \\alpha, \\tan \\alpha, \\cot \\alpha$.<\/li>\n<li><strong>Ghi ch\u00fa quan tr\u1ecdng<\/strong>:\n<ul>\n<li>B\u1ea3ng gi\u00e1 tr\u1ecb l\u01b0\u1ee3ng gi\u00e1c: $0^\\circ, 30^\\circ, 45^\\circ, 60^\\circ, 90^\\circ$.<\/li>\n<li>Quan h\u1ec7: $\\sin^2 \\alpha + \\cos^2 \\alpha = 1$, $\\tan \\alpha = \\frac{\\sin \\alpha}{\\cos \\alpha}$.<\/li>\n<\/ul>\n<\/li>\n<li><strong>C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\n<ul>\n<li>$\\sin^2 \\alpha + \\cos^2 \\alpha = 1$.<\/li>\n<li>$\\tan \\alpha \\cdot \\cot \\alpha = 1$.<\/li>\n<li>Gi\u00e1 tr\u1ecb \u0111\u1eb7c bi\u1ec7t:\n<ul>\n<li>$\\sin 30^\\circ = \\frac{1}{2}, \\cos 30^\\circ = \\frac{\\sqrt{3}}{2}$.<\/li>\n<li>$\\sin 45^\\circ = \\cos 45^\\circ = \\frac{\\sqrt{2}}{2}$.<\/li>\n<li>$\\sin 60^\\circ = \\frac{\\sqrt{3}}{2}, \\cos 60^\\circ = \\frac{1}{2}$.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-469 alignnone size-medium\" src=\"https:\/\/giapsgk.com\/chuyen-de\/wp-content\/uploads\/giapsgk-cac-cong-thuc-toan-lop-10-3.png\" alt=\"H\u00ecnh \u1ea3nh - cac-cong-thuc-toan-lop-10\" width=\"917\" height=\"452\" \/><\/p>\n<h3><strong>2. \u0110\u1ecbnh l\u00fd cosin v\u00e0 \u0111\u1ecbnh l\u00fd sin<\/strong><\/h3>\n<ul>\n<li><strong>\u0110\u1ecbnh ngh\u0129a<\/strong>:\n<ul>\n<li>\u0110\u1ecbnh l\u00fd sin: Trong tam gi\u00e1c, t\u1ef7 l\u1ec7 c\u1ea1nh v\u00e0 $\\sin$ g\u00f3c \u0111\u1ed1i di\u1ec7n b\u1eb1ng nhau.<\/li>\n<li>\u0110\u1ecbnh l\u00fd cosin: Li\u00ean h\u1ec7 gi\u1eefa c\u1ea1nh v\u00e0 g\u00f3c trong tam gi\u00e1c.<\/li>\n<\/ul>\n<\/li>\n<li><strong>Ghi ch\u00fa quan tr\u1ecdng<\/strong>:\n<ul>\n<li>D\u00f9ng \u0111\u1ec3 gi\u1ea3i tam gi\u00e1c (t\u00ednh c\u1ea1nh, g\u00f3c).<\/li>\n<\/ul>\n<\/li>\n<li><strong>C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\n<ul>\n<li>\u0110\u1ecbnh l\u00fd sin: $\\frac{a}{\\sin A} = \\frac{b}{\\sin B} = \\frac{c}{\\sin C} = 2R$, v\u1edbi $R$ l\u00e0 b\u00e1n k\u00ednh \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp.<\/li>\n<li>\u0110\u1ecbnh l\u00fd cosin: $a^2 = b^2 + c^2 &#8211; 2bc \\cos A$.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<h3><strong>3. Gi\u1ea3i tam gi\u00e1c v\u00e0 \u1ee9ng d\u1ee5ng th\u1ef1c t\u1ebf<\/strong><\/h3>\n<ul>\n<li><strong>\u0110\u1ecbnh ngh\u0129a<\/strong>: T\u00ecm c\u00e1c c\u1ea1nh, g\u00f3c c\u00f2n l\u1ea1i c\u1ee7a tam gi\u00e1c khi bi\u1ebft m\u1ed9t s\u1ed1 y\u1ebfu t\u1ed1.<\/li>\n<li><strong>Ghi ch\u00fa quan tr\u1ecdng<\/strong>:\n<ul>\n<li>\u1ee8ng d\u1ee5ng trong \u0111o \u0111\u1ea1c, t\u00ednh kho\u1ea3ng c\u00e1ch, chi\u1ec1u cao.<\/li>\n<\/ul>\n<\/li>\n<li><strong>C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\n<ul>\n<li>Di\u1ec7n t\u00edch tam gi\u00e1c: $S = \\frac{1}{2}bc \\sin A$.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<h3><strong>4. H\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong tam gi\u00e1c<\/strong><\/h3>\n<ul>\n<li><strong>\u0110\u1ecbnh ngh\u0129a<\/strong>: C\u00e1c c\u00f4ng th\u1ee9c li\u00ean h\u1ec7 gi\u1eefa c\u1ea1nh, g\u00f3c, \u0111\u01b0\u1eddng trung tuy\u1ebfn, ph\u00e2n gi\u00e1c.<\/li>\n<li><strong>Ghi ch\u00fa quan tr\u1ecdng<\/strong>:\n<ul>\n<li>C\u00f4ng th\u1ee9c di\u1ec7n t\u00edch v\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn th\u01b0\u1eddng d\u00f9ng trong b\u00e0i to\u00e1n th\u1ef1c t\u1ebf.<\/li>\n<\/ul>\n<\/li>\n<li><strong>C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\n<ul>\n<li>Di\u1ec7n t\u00edch: $S = \\frac{1}{2}ab \\sin C$.<\/li>\n<li>\u0110\u01b0\u1eddng trung tuy\u1ebfn: $m_a = \\frac{1}{2} \\sqrt{2b^2 + 2c^2 &#8211; a^2}$.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<h2><strong>Ch\u01b0\u01a1ng V: Vect\u01a1<\/strong><\/h2>\n<h3><strong>1. Kh\u00e1i ni\u1ec7m vect\u01a1<\/strong><\/h3>\n<ul>\n<li><strong>\u0110\u1ecbnh ngh\u0129a<\/strong>: Vect\u01a1 l\u00e0 \u0111\u1ea1i l\u01b0\u1ee3ng c\u00f3 \u0111\u1ed9 l\u1edbn v\u00e0 h\u01b0\u1edbng. K\u00fd hi\u1ec7u: $\\vec{v}$.<\/li>\n<li><strong>Ghi ch\u00fa quan tr\u1ecdng<\/strong>:\n<ul>\n<li>Vect\u01a1 b\u1eb1ng nhau: C\u00f9ng \u0111\u1ed9 l\u1edbn, c\u00f9ng h\u01b0\u1edbng.<\/li>\n<li>Vect\u01a1 kh\u00f4ng: $\\vec{0}$, \u0111\u1ed9 l\u1edbn b\u1eb1ng 0.<\/li>\n<\/ul>\n<\/li>\n<li><strong>C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\n<ul>\n<li>\u0110\u1ed9 l\u1edbn: $|\\vec{v}| = \\sqrt{x^2 + y^2}$ (trong m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9).<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<h3><strong>2. T\u1ed5ng v\u00e0 hi\u1ec7u c\u1ee7a hai vect\u01a1<\/strong><\/h3>\n<ul>\n<li><strong>\u0110\u1ecbnh ngh\u0129a<\/strong>: T\u1ed5ng\/hi\u1ec7u vect\u01a1 \u0111\u01b0\u1ee3c th\u1ef1c hi\u1ec7n theo quy t\u1eafc h\u00ecnh b\u00ecnh h\u00e0nh ho\u1eb7c tam gi\u00e1c.<\/li>\n<li><strong>Ghi ch\u00fa quan tr\u1ecdng<\/strong>:\n<ul>\n<li>T\u00ednh ch\u1ea5t: Giao ho\u00e1n, k\u1ebft h\u1ee3p.<\/li>\n<\/ul>\n<\/li>\n<li><strong>C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\n<ul>\n<li>T\u1ed5ng: $\\vec{a} + \\vec{b} = (x_a + x_b, y_a + y_b)$.<\/li>\n<li>Hi\u1ec7u: $\\vec{a} &#8211; \\vec{b} = (x_a &#8211; x_b, y_a &#8211; y_b)$.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<h3><strong>3. T\u00edch c\u1ee7a m\u1ed9t s\u1ed1 v\u1edbi m\u1ed9t vect\u01a1<\/strong><\/h3>\n<ul>\n<li><strong>\u0110\u1ecbnh ngh\u0129a<\/strong>: T\u00edch $k\\vec{v}$ l\u00e0 vect\u01a1 c\u00f3 \u0111\u1ed9 l\u1edbn $|k| \\cdot |\\vec{v}|$, c\u00f9ng ho\u1eb7c ng\u01b0\u1ee3c h\u01b0\u1edbng v\u1edbi $\\vec{v}$.<\/li>\n<li><strong>Ghi ch\u00fa quan tr\u1ecdng<\/strong>:\n<ul>\n<li>$k &gt; 0$: C\u00f9ng h\u01b0\u1edbng.<\/li>\n<li>$k &lt; 0$: Ng\u01b0\u1ee3c h\u01b0\u1edbng.<\/li>\n<\/ul>\n<\/li>\n<li><strong>C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\u00a0$k\\vec{v} = (kx, ky)$.<\/li>\n<\/ul>\n<h3><strong>4. Vect\u01a1 trong m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9<\/strong><\/h3>\n<ul>\n<li><strong>\u0110\u1ecbnh ngh\u0129a<\/strong>: Vect\u01a1 \u0111\u01b0\u1ee3c bi\u1ec3u di\u1ec5n b\u1eb1ng t\u1ecda \u0111\u1ed9 $\\vec{v} = (x, y)$.<\/li>\n<li><strong>Ghi ch\u00fa quan tr\u1ecdng<\/strong>:\n<ul>\n<li>T\u1ecda \u0111\u1ed9 c\u1ee7a t\u1ed5ng, hi\u1ec7u, t\u00edch \u0111\u01b0\u1ee3c t\u00ednh tr\u1ef1c ti\u1ebfp t\u1eeb t\u1ecda \u0111\u1ed9.<\/li>\n<\/ul>\n<\/li>\n<li><strong>C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\n<ul>\n<li>T\u1ecda \u0111\u1ed9: $\\vec{v} = (x_2 &#8211; x_1, y_2 &#8211; y_1)$ v\u1edbi hai \u0111i\u1ec3m $A(x_1, y_1)$, $B(x_2, y_2)$.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<h3><strong>5. T\u00edch v\u00f4 h\u01b0\u1edbng c\u1ee7a hai vect\u01a1<\/strong><\/h3>\n<ul>\n<li><strong>\u0110\u1ecbnh ngh\u0129a<\/strong>: T\u00edch v\u00f4 h\u01b0\u1edbng $\\vec{a} \\cdot \\vec{b}$ l\u00e0 s\u1ed1 th\u1ef1c, t\u00ednh b\u1eb1ng \u0111\u1ed9 l\u1edbn v\u00e0 g\u00f3c gi\u1eefa hai vect\u01a1.<\/li>\n<li><strong>Ghi ch\u00fa quan tr\u1ecdng<\/strong>:\n<ul>\n<li>Hai vect\u01a1 vu\u00f4ng g\u00f3c: $\\vec{a} \\cdot \\vec{b} = 0$.<\/li>\n<\/ul>\n<\/li>\n<li><strong>C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\n<ul>\n<li>$\\vec{a} \\cdot \\vec{b} = |\\vec{a}| |\\vec{b}| \\cos \\theta$.<\/li>\n<li>Trong t\u1ecda \u0111\u1ed9: $\\vec{a} \\cdot \\vec{b} = x_a x_b + y_a y_b$.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<h2><strong>Ch\u01b0\u01a1ng VI: Th\u1ed1ng k\u00ea<\/strong><\/h2>\n<h3><strong>1. S\u1ed1 g\u1ea7n \u0111\u00fang v\u00e0 sai s\u1ed1<\/strong><\/h3>\n<ul>\n<li><strong>\u0110\u1ecbnh ngh\u0129a<\/strong>:\n<ul>\n<li>S\u1ed1 g\u1ea7n \u0111\u00fang: Gi\u00e1 tr\u1ecb x\u1ea5p x\u1ec9 c\u1ee7a m\u1ed9t s\u1ed1.<\/li>\n<li>Sai s\u1ed1 tuy\u1ec7t \u0111\u1ed1i: $\\Delta a = |a &#8211; \\tilde{a}|$.<\/li>\n<li>Sai s\u1ed1 t\u01b0\u01a1ng \u0111\u1ed1i: $\\frac{\\Delta a}{|a|}$.<\/li>\n<\/ul>\n<\/li>\n<li><strong>Ghi ch\u00fa quan tr\u1ecdng<\/strong>:\n<ul>\n<li>D\u00f9ng trong \u0111o l\u01b0\u1eddng, t\u00ednh to\u00e1n th\u1ef1c t\u1ebf.<\/li>\n<\/ul>\n<\/li>\n<li><strong>C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\n<ul>\n<li>Sai s\u1ed1 tuy\u1ec7t \u0111\u1ed1i: $\\Delta a = |a &#8211; \\tilde{a}|$.<\/li>\n<li>Sai s\u1ed1 t\u01b0\u01a1ng \u0111\u1ed1i: $\\frac{\\Delta a}{|a|} \\cdot 100%$.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<h3><strong>2. Bi\u1ec3u di\u1ec5n d\u1eef li\u1ec7u<\/strong><\/h3>\n<ul>\n<li><strong>\u0110\u1ecbnh ngh\u0129a<\/strong>: D\u1eef li\u1ec7u \u0111\u01b0\u1ee3c bi\u1ec3u di\u1ec5n qua b\u1ea3ng t\u1ea7n s\u1ed1, bi\u1ec3u \u0111\u1ed3 c\u1ed9t, h\u00ecnh qu\u1ea1t.<\/li>\n<li><strong>Ghi ch\u00fa quan tr\u1ecdng<\/strong>:\n<ul>\n<li>Bi\u1ec3u \u0111\u1ed3 gi\u00fap tr\u1ef1c quan h\u00f3a d\u1eef li\u1ec7u.<\/li>\n<\/ul>\n<\/li>\n<li><strong>C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\n<ul>\n<li>T\u1ea7n s\u1ed1 t\u01b0\u01a1ng \u0111\u1ed1i: $f_i = \\frac{n_i}{n}$, v\u1edbi $n_i$ l\u00e0 t\u1ea7n s\u1ed1, $n$ l\u00e0 t\u1ed5ng s\u1ed1 li\u1ec7u.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<h3><strong>3. C\u00e1c s\u1ed1 \u0111\u1eb7c tr\u01b0ng \u0111o xu th\u1ebf trung t\u00e2m<\/strong><\/h3>\n<ul>\n<li><strong>\u0110\u1ecbnh ngh\u0129a<\/strong>:\n<ul>\n<li>Trung b\u00ecnh: $\\bar{x} = \\frac{\\sum x_i}{n}$.<\/li>\n<li>Trung v\u1ecb: Gi\u00e1 tr\u1ecb \u1edf gi\u1eefa khi s\u1eafp x\u1ebfp d\u1eef li\u1ec7u.<\/li>\n<li>M\u1ed1t: Gi\u00e1 tr\u1ecb xu\u1ea5t hi\u1ec7n nhi\u1ec1u nh\u1ea5t.<\/li>\n<\/ul>\n<\/li>\n<li><strong>Ghi ch\u00fa quan tr\u1ecdng<\/strong>:\n<ul>\n<li>Trung b\u00ecnh nh\u1ea1y v\u1edbi gi\u00e1 tr\u1ecb b\u1ea5t th\u01b0\u1eddng, trung v\u1ecb \u1ed5n \u0111\u1ecbnh h\u01a1n.<\/li>\n<\/ul>\n<\/li>\n<li><strong>C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\n<ul>\n<li>Trung b\u00ecnh: $\\bar{x} = \\frac{x_1 + x_2 + \\cdots + x_n}{n}$.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<h3><strong>4. C\u00e1c s\u1ed1 \u0111\u1eb7c tr\u01b0ng \u0111o \u0111\u1ed9 ph\u00e2n t\u00e1n<\/strong><\/h3>\n<ul>\n<li><strong>\u0110\u1ecbnh ngh\u0129a<\/strong>:\n<ul>\n<li>Kho\u1ea3ng bi\u1ebfn thi\u00ean: $R = x_{\\text{max}} &#8211; x_{\\text{min}}$.<\/li>\n<li>Ph\u01b0\u01a1ng sai: $\\sigma^2 = \\frac{\\sum (x_i &#8211; \\bar{x})^2}{n}$.<\/li>\n<li>\u0110\u1ed9 l\u1ec7ch chu\u1ea9n: $\\sigma = \\sqrt{\\sigma^2}$.<\/li>\n<\/ul>\n<\/li>\n<li><strong>Ghi ch\u00fa quan tr\u1ecdng<\/strong>:\n<ul>\n<li>\u0110\u1ed9 l\u1ec7ch chu\u1ea9n c\u00e0ng l\u1edbn, d\u1eef li\u1ec7u c\u00e0ng ph\u00e2n t\u00e1n.<\/li>\n<\/ul>\n<\/li>\n<li><strong>C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\n<ul>\n<li>Ph\u01b0\u01a1ng sai: $\\sigma^2 = \\frac{\\sum x_i^2}{n} &#8211; \\bar{x}^2$.<\/li>\n<li>\u0110\u1ed9 l\u1ec7ch chu\u1ea9n: $\\sigma = \\sqrt{\\frac{\\sum (x_i &#8211; \\bar{x})^2}{n}}$.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<h2><strong>Ch\u01b0\u01a1ng VII: \u0110\u1ea1i s\u1ed1 t\u1ed5 h\u1ee3p<\/strong><\/h2>\n<h3><strong>1. Quy t\u1eafc \u0111\u1ebfm<\/strong><\/h3>\n<ul>\n<li><strong>\u0110\u1ecbnh ngh\u0129a<\/strong>:\n<ul>\n<li>Quy t\u1eafc c\u1ed9ng: S\u1ed1 c\u00e1ch th\u1ef1c hi\u1ec7n m\u1ed9t trong hai h\u00e0nh \u0111\u1ed9ng.<\/li>\n<li>Quy t\u1eafc nh\u00e2n: S\u1ed1 c\u00e1ch th\u1ef1c hi\u1ec7n tu\u1ea7n t\u1ef1 c\u00e1c h\u00e0nh \u0111\u1ed9ng.<\/li>\n<\/ul>\n<\/li>\n<li><strong>Ghi ch\u00fa quan tr\u1ecdng<\/strong>:\n<ul>\n<li>S\u01a1 \u0111\u1ed3 h\u00ecnh c\u00e2y gi\u00fap li\u1ec7t k\u00ea c\u00e1c tr\u01b0\u1eddng h\u1ee3p.<\/li>\n<\/ul>\n<\/li>\n<li><strong>C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\n<ul>\n<li>Quy t\u1eafc nh\u00e2n: N\u1ebfu h\u00e0nh \u0111\u1ed9ng 1 c\u00f3 $m$ c\u00e1ch, h\u00e0nh \u0111\u1ed9ng 2 c\u00f3 $n$ c\u00e1ch, th\u00ec t\u1ed5ng s\u1ed1 c\u00e1ch l\u00e0 $m \\cdot n$.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<h3><strong>2. Ho\u00e1n v\u1ecb, ch\u1ec9nh h\u1ee3p, t\u1ed5 h\u1ee3p<\/strong><\/h3>\n<ul>\n<li><strong>\u0110\u1ecbnh ngh\u0129a<\/strong>:\n<ul>\n<li>Ho\u00e1n v\u1ecb: S\u1eafp x\u1ebfp $n$ ph\u1ea7n t\u1eed, $P_n = n!$.<\/li>\n<li>Ch\u1ec9nh h\u1ee3p: Ch\u1ecdn v\u00e0 s\u1eafp x\u1ebfp $k$ ph\u1ea7n t\u1eed t\u1eeb $n$, $A_n^k = \\frac{n!}{(n-k)!}$.<\/li>\n<li>T\u1ed5 h\u1ee3p: Ch\u1ecdn $k$ ph\u1ea7n t\u1eed t\u1eeb $n$, kh\u00f4ng quan t\u00e2m th\u1ee9 t\u1ef1, $C_n^k = \\frac{n!}{k!(n-k)!}$.<\/li>\n<\/ul>\n<\/li>\n<li><strong>Ghi ch\u00fa quan tr\u1ecdng<\/strong>:\n<ul>\n<li>$C_n^k = C_n^{n-k}$.<\/li>\n<\/ul>\n<\/li>\n<li><strong>C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\n<ul>\n<li>$P_n = n!$.<\/li>\n<li>$A_n^k = n \\cdot (n-1) \\cdot \\ldots \\cdot (n-k+1)$.<\/li>\n<li>$C_n^k = \\frac{A_n^k}{k!}$.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<h3><strong>3. Nh\u1ecb th\u1ee9c Newton<\/strong><\/h3>\n<ul>\n<li><strong>\u0110\u1ecbnh ngh\u0129a<\/strong>: Khai tri\u1ec3n $(a + b)^n$ th\u00e0nh t\u1ed5ng c\u00e1c h\u1ea1ng.<\/li>\n<li><strong>Ghi ch\u00fa quan tr\u1ecdng<\/strong>:\n<ul>\n<li>H\u1ec7 s\u1ed1 $C_n^k$ \u0111\u01b0\u1ee3c g\u1ecdi l\u00e0 h\u1ec7 s\u1ed1 nh\u1ecb th\u1ee9c.<\/li>\n<\/ul>\n<\/li>\n<li><strong>C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\n<ul>\n<li>$(a + b)^n = \\sum_{k=0}^n C_n^k a^{n-k} b^k$.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<h2><strong>Ch\u01b0\u01a1ng VIII: X\u00e1c su\u1ea5t<\/strong><\/h2>\n<h3><strong>1. Bi\u1ebfn c\u1ed1 v\u00e0 Kh\u00f4ng gian m\u1eabu<\/strong><\/h3>\n<ul>\n<li><strong>\u0110\u1ecbnh ngh\u0129a<\/strong>:\n<ul>\n<li>Kh\u00f4ng gian m\u1eabu ($\\Omega$): T\u1eadp h\u1ee3p t\u1ea5t c\u1ea3 k\u1ebft qu\u1ea3 c\u00f3 th\u1ec3 c\u1ee7a th\u00ed nghi\u1ec7m.<\/li>\n<li>Bi\u1ebfn c\u1ed1: T\u1eadp con c\u1ee7a kh\u00f4ng gian m\u1eabu.<\/li>\n<\/ul>\n<\/li>\n<li><strong>Ghi ch\u00fa quan tr\u1ecdng<\/strong>:\n<ul>\n<li>Bi\u1ebfn c\u1ed1 b\u1ea5t kh\u1ea3: $\\emptyset$, bi\u1ebfn c\u1ed1 ch\u1eafc ch\u1eafn: $\\Omega$.<\/li>\n<\/ul>\n<\/li>\n<li><strong>C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\n<ul>\n<li>S\u1ed1 ph\u1ea7n t\u1eed: $|A| \\leq |\\Omega|$.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<h3><strong>2. X\u00e1c su\u1ea5t c\u1ee7a bi\u1ebfn c\u1ed1<\/strong><\/h3>\n<ul>\n<li><strong>\u0110\u1ecbnh ngh\u0129a<\/strong>: X\u00e1c su\u1ea5t c\u1ed5 \u0111i\u1ec3n $P(A) = \\frac{|A|}{|\\Omega|}$.<\/li>\n<li><strong>Ghi ch\u00fa quan tr\u1ecdng<\/strong>:\n<ul>\n<li>$0 \\leq P(A) \\leq 1$.<\/li>\n<\/ul>\n<\/li>\n<li><strong>C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\n<ul>\n<li>$P(A) = \\frac{\\text{s\u1ed1 tr\u01b0\u1eddng h\u1ee3p thu\u1eadn l\u1ee3i}}{\\text{t\u1ed5ng s\u1ed1 tr\u01b0\u1eddng h\u1ee3p}}$.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<h2><strong>Ch\u01b0\u01a1ng IX: Ph\u01b0\u01a1ng ph\u00e1p t\u1ecda \u0111\u1ed9 trong m\u1eb7t ph\u1eb3ng<\/strong><\/h2>\n<h3><strong>1. Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng<\/strong><\/h3>\n<ul>\n<li><strong>\u0110\u1ecbnh ngh\u0129a<\/strong>: Ph\u01b0\u01a1ng tr\u00ecnh bi\u1ec3u di\u1ec5n \u0111\u01b0\u1eddng th\u1eb3ng tr\u00ean m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9.<\/li>\n<li><strong>Ghi ch\u00fa quan tr\u1ecdng<\/strong>:\n<ul>\n<li>C\u00e1c d\u1ea1ng: T\u1ed5ng qu\u00e1t, tham s\u1ed1, ch\u00ednh t\u1eafc.<\/li>\n<\/ul>\n<\/li>\n<li><strong>C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\n<ul>\n<li>T\u1ed5ng qu\u00e1t: $ax + by + c = 0$.<\/li>\n<li>Ch\u00ednh t\u1eafc: $\\frac{x &#8211; x_0}{a} = \\frac{y &#8211; y_0}{b}$.<\/li>\n<li>Tham s\u1ed1: $\\begin{cases} x = x_0 + at \\ y = y_0 + bt \\end{cases}$.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<h3><strong>2. V\u1ecb tr\u00ed t\u01b0\u01a1ng \u0111\u1ed1i, g\u00f3c, kho\u1ea3ng c\u00e1ch<\/strong><\/h3>\n<ul>\n<li><strong>\u0110\u1ecbnh ngh\u0129a<\/strong>:\n<ul>\n<li>V\u1ecb tr\u00ed t\u01b0\u01a1ng \u0111\u1ed1i: Song song, c\u1eaft nhau, vu\u00f4ng g\u00f3c.<\/li>\n<li>G\u00f3c: G\u00f3c gi\u1eefa hai \u0111\u01b0\u1eddng th\u1eb3ng.<\/li>\n<li>Kho\u1ea3ng c\u00e1ch: T\u1eeb \u0111i\u1ec3m \u0111\u1ebfn \u0111\u01b0\u1eddng th\u1eb3ng.<\/li>\n<\/ul>\n<\/li>\n<li><strong>Ghi ch\u00fa quan tr\u1ecdng<\/strong>:\n<ul>\n<li>Hai \u0111\u01b0\u1eddng th\u1eb3ng vu\u00f4ng g\u00f3c: T\u00edch c\u00e1c vect\u01a1 ch\u1ec9 ph\u01b0\u01a1ng b\u1eb1ng 0.<\/li>\n<\/ul>\n<\/li>\n<li><strong>C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\n<ul>\n<li>Kho\u1ea3ng c\u00e1ch t\u1eeb \u0111i\u1ec3m $M(x_0, y_0)$ \u0111\u1ebfn \u0111\u01b0\u1eddng th\u1eb3ng $ax + by + c = 0$: $d = \\frac{|ax_0 + by_0 + c|}{\\sqrt{a^2 + b^2}}$.<\/li>\n<li>G\u00f3c gi\u1eefa hai \u0111\u01b0\u1eddng th\u1eb3ng: $\\cos \\theta = \\frac{|\\vec{u} \\cdot \\vec{v}|}{|\\vec{u}| |\\vec{v}|}$.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<h3><strong>3. Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng tr\u00f2n<\/strong><\/h3>\n<ul>\n<li><strong>\u0110\u1ecbnh ngh\u0129a<\/strong>: T\u1eadp h\u1ee3p c\u00e1c \u0111i\u1ec3m c\u00e1ch t\u00e2m m\u1ed9t kho\u1ea3ng c\u1ed1 \u0111\u1ecbnh.<\/li>\n<li><strong>Ghi ch\u00fa quan tr\u1ecdng<\/strong>:\n<ul>\n<li>D\u1ea1ng t\u1ed5ng qu\u00e1t c\u00f3 th\u1ec3 chuy\u1ec3n th\u00e0nh d\u1ea1ng chu\u1ea9n.<\/li>\n<\/ul>\n<\/li>\n<li><strong>C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\n<ul>\n<li>D\u1ea1ng chu\u1ea9n: $(x &#8211; a)^2 + (y &#8211; b)^2 = R^2$.<\/li>\n<li>D\u1ea1ng t\u1ed5ng qu\u00e1t: $x^2 + y^2 + Dx + Ey + F = 0$.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<h3><strong>4. Ba \u0111\u01b0\u1eddng conic<\/strong><\/h3>\n<ul>\n<li><strong>\u0110\u1ecbnh ngh\u0129a<\/strong>:\n<ul>\n<li>Elip: $\\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1$.<\/li>\n<li>Hyperbol: $\\frac{x^2}{a^2} &#8211; \\frac{y^2}{b^2} = 1$.<\/li>\n<li>Parabol: $y^2 = 2px$.<\/li>\n<\/ul>\n<\/li>\n<li><strong>Ghi ch\u00fa quan tr\u1ecdng<\/strong>:\n<ul>\n<li>Ph\u00e2n bi\u1ec7t d\u1ef1a v\u00e0o d\u1ea1ng ph\u01b0\u01a1ng tr\u00ecnh.<\/li>\n<\/ul>\n<\/li>\n<li><strong>C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\n<ul>\n<li>Elip: $\\frac{(x &#8211; h)^2}{a^2} + \\frac{(y &#8211; k)^2}{b^2} = 1$.<\/li>\n<li>Hyperbol: $\\frac{(x &#8211; h)^2}{a^2} &#8211; \\frac{(y &#8211; k)^2}{b^2} = 1$.<\/li>\n<li>Parabol: $y^2 = 4ax$.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<h2><strong>Ho\u1ea1t \u0111\u1ed9ng th\u1ef1c h\u00e0nh v\u00e0 tr\u1ea3i nghi\u1ec7m<\/strong><\/h2>\n<h3><strong>\u1ee8ng d\u1ee5ng th\u1ef1c t\u1ebf<\/strong><\/h3>\n<ul>\n<li><strong>T\u00e0i ch\u00ednh<\/strong>: T\u00ednh l\u00e3i su\u1ea5t, chi ph\u00ed.\n<ul>\n<li>C\u00f4ng th\u1ee9c: L\u00e3i \u0111\u01a1n $I = P \\cdot r \\cdot t$, l\u00e3i k\u00e9p $A = P \\left(1 + \\frac{r}{n}\\right)^{nt}$.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p style=\"text-align: center;\"><a href=\"https:\/\/giapsgk.com\/chuyen-de\/wp-content\/uploads\/Tom-tat-kien-thuc-TOAN-HOC-lop-10.pdf\">File pdf T\u00f3m t\u1eaft ki\u1ebfn th\u1ee9c TO\u00c1N H\u1eccC l\u1edbp 10<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"<p>To\u00e1n h\u1ecdc l\u1edbp 10 l\u00e0 b\u01b0\u1edbc kh\u1edfi \u0111\u1ea7u quan tr\u1ecdng trong ch\u01b0\u01a1ng tr\u00ecnh trung h\u1ecdc ph\u1ed5 th\u00f4ng, n\u01a1i h\u1ecdc sinh l\u00e0m quen v\u1edbi nh\u1eefng kh\u00e1i ni\u1ec7m tr\u1eebu t\u01b0\u1ee3ng h\u01a1n v\u00e0 t\u01b0 duy logic s\u00e2u s\u1eafc h\u01a1n. Vi\u1ec7c n\u1eafm ch\u1eafc c\u00e1c c\u00f4ng th\u1ee9c to\u00e1n h\u1ecdc l\u1edbp 10 kh\u00f4ng ch\u1ec9 gi\u00fap c\u00e1c em h\u1ecdc t\u1ed1t hi\u1ec7n t\u1ea1i, &#8230; <a title=\"H\u1ec7 th\u1ed1ng c\u00e1c c\u00f4ng th\u1ee9c To\u00e1n 10 (h\u1ecdc k\u00ec 1 v\u00e0 h\u1ecdc k\u00ec 2 &#8211; s\u00e1ch m\u1edbi)\" class=\"read-more\" href=\"https:\/\/giapsgk.com\/chuyen-de\/cac-cong-thuc-toan-lop-10\/\" aria-label=\"\u0110\u1ecdc th\u00eam v\u1ec1 H\u1ec7 th\u1ed1ng c\u00e1c c\u00f4ng th\u1ee9c To\u00e1n 10 (h\u1ecdc k\u00ec 1 v\u00e0 h\u1ecdc k\u00ec 2 &#8211; s\u00e1ch m\u1edbi)\">\u0110\u1ecdc ti\u1ebfp<\/a><\/p>\n","protected":false},"author":1,"featured_media":466,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[4],"tags":[],"class_list":["post-458","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-ly-thuyet"],"_links":{"self":[{"href":"https:\/\/giapsgk.com\/chuyen-de\/wp-json\/wp\/v2\/posts\/458","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/giapsgk.com\/chuyen-de\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/giapsgk.com\/chuyen-de\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/giapsgk.com\/chuyen-de\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/giapsgk.com\/chuyen-de\/wp-json\/wp\/v2\/comments?post=458"}],"version-history":[{"count":10,"href":"https:\/\/giapsgk.com\/chuyen-de\/wp-json\/wp\/v2\/posts\/458\/revisions"}],"predecessor-version":[{"id":472,"href":"https:\/\/giapsgk.com\/chuyen-de\/wp-json\/wp\/v2\/posts\/458\/revisions\/472"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/giapsgk.com\/chuyen-de\/wp-json\/wp\/v2\/media\/466"}],"wp:attachment":[{"href":"https:\/\/giapsgk.com\/chuyen-de\/wp-json\/wp\/v2\/media?parent=458"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/giapsgk.com\/chuyen-de\/wp-json\/wp\/v2\/categories?post=458"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/giapsgk.com\/chuyen-de\/wp-json\/wp\/v2\/tags?post=458"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}