{"segment":[{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<span class='basic_left'>T\u1ee9 gi\u00e1c c\u00f3 hai \u0111\u01b0\u1eddng ch\u00e9o vu\u00f4ng g\u00f3c v\u1edbi nhau, trong \u0111\u00f3 m\u1ed9t \u0111\u01b0\u1eddng ch\u00e9o c\u00f3 \u0111\u1ed9 d\u00e0i $x$ v\u00e0 \u0111\u01b0\u1eddng ch\u00e9o c\u00f2n l\u1ea1i g\u1ea5p r\u01b0\u1ee1i \u0111\u01b0\u1eddng ch\u00e9o kia. H\u00e0m s\u1ed1 bi\u1ec3u di\u1ec5n di\u1ec7n t\u00edch c\u1ee7a t\u1ee9 gi\u00e1c \u0111\u00f3 l\u00e0 $y=\\dfrac{1}{4}x^2,$ <b>\u0111\u00fang<\/b> hay <b> sai<\/b>?<\/span>","select":["A. \u0110\u00fang","B. Sai"],"hint":"Di\u1ec7n t\u00edch t\u1ee9 gi\u00e1c c\u00f3 hai \u0111\u01b0\u1eddng ch\u00e9o vu\u00f4ng g\u00f3c b\u1eb1ng n\u1eeda t\u00edch hai \u0111\u01b0\u1eddng ch\u00e9o \u0111\u00f3.","explain":"<span class='basic_left'>T\u1ee9 gi\u00e1c c\u00f3 m\u1ed9t \u0111\u01b0\u1eddng ch\u00e9o c\u00f3 \u0111\u1ed9 d\u00e0i $x;$ \u0111\u01b0\u1eddng ch\u00e9o c\u00f2n l\u1ea1i g\u1ea5p r\u01b0\u1ee1i \u0111\u01b0\u1eddng ch\u00e9o n\u00e0y n\u00ean c\u00f3 \u0111\u1ed9 d\u00e0i $\\dfrac{3}{2}x$ <br\/>Suy ra di\u1ec7n t\u00edch t\u1ee9 gi\u00e1c l\u00e0 $y=\\dfrac{1}{2}.x.\\dfrac{3}{2}x\\Leftrightarrow y=\\dfrac{3}{4}{{x}^{2}}$ <br\/>V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 sai<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 B<\/span><\/span>","column":2}]}],"id_ques":791},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\dfrac{6}{25};\\dfrac{3}{2};24;54;\\dfrac{27}{2}$","B. $\\dfrac{6}{25};\\dfrac{3}{2};54;24;\\dfrac{27}{2}$","C. $\\dfrac{6}{25};\\dfrac{3}{2};14;54;\\dfrac{27}{2}$"],"ques":"<span class='basic_left'>Cho h\u00ecnh l\u1eadp ph\u01b0\u01a1ng c\u00f3 \u0111\u1ed9 d\u00e0i c\u1ea1nh l\u00e0 $x$ v\u00e0 di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n $S.$ <br\/><b> C\u00e2u 1:<\/b>T\u00ednh c\u00e1c gi\u00e1 tr\u1ecb c\u1ee7a $S$ \u1ee9ng v\u1edbi c\u00e1c gi\u00e1 tr\u1ecb c\u1ee7a $x$ cho trong b\u1ea3ng d\u01b0\u1edbi \u0111\u00e2y r\u1ed3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng:<br\/><table> <tr> <td>$x$<\/td> <td>$\\dfrac{1}{5}$ <\/td> <td>$\\dfrac{1}{2}$<\/td><td>$2$<\/td><td>$3$<\/td> <td>$\\dfrac{3}{2}$ <\/td><\/tr> <tr><td>$S$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)<\/td><td>?<\/td><td>?<\/td> <td>?<\/td><td>?<\/td><td>?<\/td><\/tr> <\/table><\/span>","hint":"Bi\u1ec3u di\u1ec5n di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n $S$ (t\u1ee9c t\u1ed5ng di\u1ec7n t\u00edch c\u1ee7a 6 m\u1eb7t) c\u1ee7a h\u00ecnh l\u1eadp ph\u01b0\u01a1ng qua $x$","explain":"<span class='basic_left'>Di\u1ec7n t\u00edch c\u1ee7a m\u1ed9t m\u1eb7t l\u00e0 ${{x}^{2}}$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)<br\/>Di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n c\u1ee7a h\u00ecnh l\u1eadp ph\u01b0\u01a1ng l\u00e0 $S=6{{x}^{2}}$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)<br\/>Thay l\u1ea7n l\u01b0\u1ee3t c\u00e1c gi\u00e1 tr\u1ecb c\u1ee7a $x$ l\u00e0: $\\dfrac{1}{5}$; $\\dfrac{1}{2}$;$2$; $3$;$\\dfrac{3}{2}$ v\u00e0o $S=6{{x}^{2}}$, ta c\u00f3 k\u1ebft qu\u1ea3 trong b\u1ea3ng sau: <br\/><table> <tr> <td>$x$<\/td> <td>$\\dfrac{1}{5}$ <\/td> <td>$\\dfrac{1}{2}$<\/td><td>$2$<\/td><td>$3$<\/td> <td>$\\dfrac{3}{2}$ <\/td><\/tr> <tr><td>$S$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)<\/td><td>$\\dfrac{6}{25}$ <\/td><td>$\\dfrac{3}{2}$ <\/td> <td>$24$<\/td><td>$54$<\/td><td>$\\dfrac{27}{2}$ <\/td><\/tr> <\/table><\/span>"}]}],"id_ques":792},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<span class='basic_left'>Cho h\u00ecnh l\u1eadp ph\u01b0\u01a1ng c\u00f3 \u0111\u1ed9 d\u00e0i c\u1ea1nh l\u00e0 $x$ v\u00e0 di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n $S.$<br\/> <b> C\u00e2u 2:<\/b> Khi di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n gi\u1ea3m $25$ l\u1ea7n th\u00ec c\u1ea1nh $x$ gi\u1ea3m \u0111i $25$ l\u1ea7n, <b>\u0111\u00fang<\/b> hay <b> sai<\/b>?<\/span> ","select":["A. \u0110\u00fang","B. Sai"],"hint":"","explain":"<span class='basic_left'>Di\u1ec7n t\u00edch c\u1ee7a m\u1ed9t m\u1eb7t l\u00e0 ${{x}^{2}}$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)<br\/>Di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n c\u1ee7a h\u00ecnh l\u1eadp ph\u01b0\u01a1ng l\u00e0 $S=6{{x}^{2}}$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)<br\/>Khi di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n gi\u1ea3m \u0111i $25$ l\u1ea7n, gi\u00e1 tr\u1ecb c\u1ee7a n\u00f3 l\u00e0 $S\u2019$ v\u00e0 c\u1ea1nh h\u00ecnh l\u1eadp ph\u01b0\u01a1ng khi \u0111\u00f3 l\u00e0 $x\u2019$<br\/>Ta c\u00f3 $S'=\\dfrac{S}{25}\\Leftrightarrow 6x{{'}^{2}}=\\dfrac{6{{x}^{2}}}{25}\\Leftrightarrow x{{'}^{2}}=\\dfrac{{{x}^{2}}}{25}\\Leftrightarrow x'=\\dfrac{x}{5}$ <br\/>V\u1eady $x$ gi\u1ea3m \u0111i $5$ l\u1ea7n. <br\/>Do \u0111\u00f3 kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 sai.<br\/><span class='basic_pink'>V\u1eady ch\u1ecdn \u0111\u00e1p \u00e1n B<\/span><\/span>","column":2}]}],"id_ques":793},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"<span class='basic_left'>Cho h\u00e0m s\u1ed1 $y=-{{x}^{2}}$ . Nh\u1eefng \u0111i\u1ec3m tr\u00ean \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 c\u00f3 tung \u0111\u1ed9 g\u1ea5p \u0111\u00f4i ho\u00e0nh \u0111\u1ed9 l\u00e0<\/span> ","select":["A. $O(0;0)$ v\u00e0 $M(-3;-6)$","B. $O(0;0)$ v\u00e0 $M(-1;-2)$","C. $O(1;2)$ v\u00e0 $M(-2;-4)$","D. $O(0;0)$ v\u00e0 $M(-2;-4)$"],"hint":"\u0110i\u1ec3m c\u00f3 tung \u0111\u1ed9 g\u1ea5p \u0111\u00f4i ho\u00e0nh \u0111\u1ed9 thu\u1ed9c \u0111\u01b0\u1eddng th\u1eb3ng $y=2x$","explain":"<span class='basic_left'>T\u1eadp c\u00e1c \u0111i\u1ec3m $M$ c\u00f3 tung \u0111\u1ed9 g\u1ea5p \u0111\u00f4i ho\u00e0nh \u0111\u1ed9 l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng $y=2x$<br\/>$M$ n\u1eb1m tr\u00ean \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y=-{{x}^{2}}$ n\u00ean t\u1ecda \u0111\u1ed9 c\u1ee7a $M$ l\u00e0 nghi\u1ec7m c\u1ee7a h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh<br\/> $\\left\\{ \\begin{aligned} & y=-{{x}^{2}} \\\\ & y=2x \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & -{{x}^{2}}=2x \\\\ & y=2x \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & \\left[ \\begin{aligned} & x=0 \\\\ & x=-2 \\\\ \\end{aligned} \\right. \\\\ & y=2x \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left[ \\begin{aligned} & x=y=0 \\\\ & x=-2;y=-4 \\\\ \\end{aligned} \\right.$ <br\/>V\u1eady c\u00f3 hai \u0111i\u1ec3m M th\u1ecfa m\u00e3n y\u00eau c\u1ea7u \u0111\u1ec1 b\u00e0i: $O(0;0)$ v\u00e0 $M(-2;-4)$<br\/><span class='basic_pink'>V\u1eady ch\u1ecdn \u0111\u00e1p \u00e1n D<\/span><\/span>","column":2}]}],"id_ques":794},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"<span class='basic_left'>Parabol trong h\u00ecnh v\u1ebd sau l\u00e0 \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 n\u00e0o?<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai19/lv2/img\/TB6.1.png' \/><\/center><\/span> ","select":["A. $y=4{{x}^{2}}$","B. $y=\\dfrac{{{x}^{2}}}{2}$ ","C. $y=\\dfrac{{{x}^{2}}}{4}$ ","D. $y=-\\dfrac{{{x}^{2}}}{4}$ "],"hint":"Parabol tr\u00ean h\u00ecnh v\u1ebd c\u00f3 d\u1ea1ng $y=a{{x}^{2}}$. T\u1eeb hai \u0111i\u1ec3m thu\u1ed9c parabol, ta t\u00ecm $a$","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai19/lv2/img\/TB6.1.png' \/><\/center>Parabol c\u1ea7n t\u00ecm c\u00f3 d\u1ea1ng $y=a{{x}^{2}}$ (P) v\u1edbi $a \\ne 0$ <br\/>Tr\u00ean h\u00ecnh v\u1ebd, ta th\u1ea5y parabol (P) \u0111i qua \u0111i\u1ec3m $(2;1)$ n\u00ean ta thay $x=2$ v\u00e0 $y=1$ v\u00e0o (P), ta c\u00f3:<br\/>$1=a.2^2\\Leftrightarrow a=\\dfrac{1}{4}$ <br\/>V\u1eady parabol trong h\u00ecnh v\u1ebd h\u00ecnh tr\u00ean l\u00e0 \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 $y=\\dfrac{{{x}^{2}}}{4}$ <br\/><span class='basic_pink'>V\u1eady ch\u1ecdn \u0111\u00e1p \u00e1n C<\/span><\/span>","column":2}]}],"id_ques":795},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["1"],["5"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai19/lv2/img\/1.png' \/><\/center>\u0110\u1ec3 \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y=\\left( {{m}^{2}}-6m \\right){{x}^{2}}$ \u0111i qua \u0111i\u1ec3m $A(1;-5)$ th\u00ec $m=$_input_ ho\u1eb7c $m=$_input_","hint":"","explain":"<span class='basic_left'>\u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y=\\left( {{m}^{2}}-6m \\right){{x}^{2}}$ \u0111i qua \u0111i\u1ec3m $A(1;-5)$ n\u00ean ta thay t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m A v\u00e0o h\u00e0m s\u1ed1:<br\/>$\\begin{aligned} & -5=\\left( {{m}^{2}}-6m \\right){{.1}^{2}}\\Leftrightarrow {{m}^{2}}-6m+5=0\\Leftrightarrow \\left( m-1 \\right)\\left( m-5 \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & m-1=0 \\\\ & m-5=0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left[ \\begin{aligned} & m=1 \\\\ & m=5 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/>V\u1eady $m=1$ ho\u1eb7c $m=5$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $1$ v\u00e0 $5$<\/span><\/span>"}]}],"id_ques":796},{"time":24,"part":[{"title":"\u0110i\u1ec1n c\u00e1c s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["1"],["2"],["3"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"<span class='basic_left'>Cho h\u00e0m s\u1ed1 $y=f\\left( x \\right)=-\\dfrac{2}{3}{{x}^{2}}$ . T\u00ecm gi\u00e1 tr\u1ecb nguy\u00ean d\u01b0\u01a1ng c\u1ee7a $m$ \u0111\u1ec3 $f\\left( m \\right)\\ge -6$<br\/><b>\u0110\u00e1p s\u1ed1: <\/b> $m \\in$ {_input_;_input_;_input_}","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: T\u00ednh $f(m)$ v\u00e0 gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh: $f\\left( m \\right)\\ge -6$<br\/>B\u01b0\u1edbc 2: T\u00ecm c\u00e1c gi\u00e1 tr\u1ecb nguy\u00ean d\u01b0\u01a1ng c\u1ee7a $m$ t\u1eeb t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh v\u1eeba gi\u1ea3i \u1edf b\u01b0\u1edbc 1<br\/><b>Ch\u00fa \u00fd: <\/b> V\u1edbi $a \\ge 0$ th\u00ec $|x|\\le a \\Leftrightarrow -a\\le x \\le a$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span> <br\/>Ta c\u00f3:<br\/>$f\\left( m \\right)=-\\dfrac{2}{3}{{m}^{2}}$ <br\/>$f\\left( m \\right)\\ge -6\\Leftrightarrow -\\dfrac{2}{3}{{m}^{2}}\\ge -6\\Leftrightarrow \\dfrac{2}{3}{{m}^{2}}\\le 6$<br\/>$\\Leftrightarrow {{m}^{2}}\\le 9\\Leftrightarrow \\left| m \\right|\\le 3\\Leftrightarrow -3\\le m\\le 3$<br\/>V\u00ec $m$ nguy\u00ean d\u01b0\u01a1ng n\u00ean $m\\in \\text{ }\\!\\!\\{\\!\\!\\text{ 1;2;3}\\}$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $1;2;3$<\/span><\/span>"}]}],"id_ques":797},{"time":24,"part":[{"title":"\u0110i\u1ec1n c\u00e1c s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["5"],["6"],["7"],["8"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"<span class='basic_left'>Cho h\u00e0m s\u1ed1 $y=f\\left( x \\right)=\\left( \\sqrt{m-5}-2 \\right){{x}^{2}}$ . T\u00ecm gi\u00e1 tr\u1ecb nguy\u00ean c\u1ee7a $m$ \u0111\u1ec3 h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn v\u1edbi $x<0$<br\/><b>\u0110\u00e1p s\u1ed1: <\/b> $m \\in$ {_input_;_input_;_input_;_input_}","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>\u0110i\u1ec1u ki\u1ec7n \u0111\u1ec3 h\u00e0m s\u1ed1 $y=a{{x}^{2}}$ \u0111\u1ed3ng bi\u1ebfn khi $x<0$ l\u00e0 $a<0$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span> <br\/>H\u00e0m s\u1ed1 $y=f\\left( x \\right)=\\left( \\sqrt{m-5}-2 \\right){{x}^{2}}$ c\u00f3 d\u1ea1ng $y=a{{x}^{2}}$ <br\/>\u0110\u1ec3 h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn v\u1edbi $x<0$ th\u00ec $a<0\\Leftrightarrow \\sqrt{m-5}-2<0\\Leftrightarrow \\sqrt{m-5}<2$ <br\/>$\\Leftrightarrow \\left\\{ \\begin{aligned} & m-5\\ge 0 \\\\ & m-5<4 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m\\ge 5 \\\\ & m<9 \\\\ \\end{aligned} \\right.\\Leftrightarrow 5\\le m<9$<br\/>Do $m$ nguy\u00ean n\u00ean $m\\in \\text{ }\\!\\!\\{\\!\\!\\text{ }5;6;7;8\\}$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $5;6;7;8.$<\/span><\/span>"}]}],"id_ques":798},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","title_trans":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","temp":"true_false","correct":[["f","f","t"]],"list":[{"point":5,"image":"img\/1.png","col_name":["","\u0110\u00fang","Sai"],"arr_ques":["1. H\u00e0m s\u1ed1 $y=\\left( {{m}^{2}}-6m+12 \\right){{x}^{2}}$ \u0111\u1ed3ng bi\u1ebfn trong kho\u1ea3ng $(-2005;0)$ v\u00e0 ngh\u1ecbch bi\u1ebfn trong kho\u1ea3ng $(0;2005)$ ","2. H\u00e0m s\u1ed1 $y=\\left( {{m}^{2}}+1 \\right){{x}^{2}}$ \u0111\u1ed3ng bi\u1ebfn tr\u00ean $\\mathbb{R}$","3. H\u00e0m s\u1ed1 $y=-\\left( {{m}^{2}}+3 \\right){{x}^{2}}$ ngh\u1ecbch bi\u1ebfn khi $x>0$ v\u00e0 \u0111\u1ed3ng bi\u1ebfn khi $x<0$"],"hint":"D\u1ef1a v\u00e0o t\u00ednh ch\u1ea5t \u0111\u1ed3ng bi\u1ebfn v\u00e0 ngh\u1ecbch bi\u1ebfn c\u1ee7a h\u00e0m s\u1ed1 $y=a x^2$ ($a \\ne 0$","explain":["<span class='basic_left'>1. Sai v\u00ec h\u00e0m s\u1ed1 $y=\\left( {{m}^{2}}-6m+12 \\right){{x}^{2}}$ c\u00f3 $a={{m}^{2}}-6m+12={{\\left( m-3 \\right)}^{2}}+3>0$ v\u1edbi m\u1ecdi $m$ n\u00ean h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn khi $x>0$ v\u00e0 ngh\u1ecbch bi\u1ebfn khi $x<0. $ Do \u0111\u00f3 h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn trong kho\u1ea3ng $(0;2005)$ v\u00e0 ngh\u1ecbch bi\u1ebfn trong kho\u1ea3ng $(-2005;0)$<\/span>","<span class='basic_left'><br\/>2. Sai v\u00ec h\u00e0m s\u1ed1 $y=\\left( {{m}^{2}}+1 \\right){{x}^{2}}$ c\u00f3 $a={{m}^{2}}+1>0$ v\u1edbi m\u1ecdi $m$ n\u00ean h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn khi $x>0$ v\u00e0 ngh\u1ecbch bi\u1ebfn khi $x<0$<\/span>","<span class='basic_left'><br\/>3. \u0110\u00fang v\u00ec h\u00e0m s\u1ed1 $y=-\\left( {{m}^{2}}+3 \\right){{x}^{2}}$c\u00f3 $a=-\\left( {{m}^{2}}+3 \\right)<0$ v\u1edbi m\u1ecdi $m$ n\u00ean h\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn khi $x>0$ v\u00e0 \u0111\u1ed3ng bi\u1ebfn khi $x<0$<\/span>"]}]}],"id_ques":799},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $m= \\frac{4}{3}; n = \\sqrt{3}$","B. $m= \\frac{1}{3}; n = \\sqrt{3}$","C. $m= \\frac{2}{3}; n = 2\\sqrt{3}$"],"ques":"<span class='basic_left'>Cho h\u00e0m s\u1ed1 $y=a{{x}^{2}}$ c\u00f3 \u0111\u1ed3 th\u1ecb \u0111i qua \u0111i\u1ec3m $A(3;3).$ T\u00ecm gi\u00e1 tr\u1ecb c\u1ee7a $m$ v\u00e0 $n$ \u0111\u1ec3 c\u00e1c \u0111i\u1ec3m $B(-2;m)$ v\u00e0 $C(n;1)$ thu\u1ed9c parabol n\u00f3i tr\u00ean v\u00e0 $n>m$<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> $m=$?$n=$?<\/span>","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: T\u00ecm $a$ t\u1eeb gi\u1ea3 thi\u1ebft \u0111\u1ed3 th\u1ecb \u0111i qua \u0111i\u1ec3m $A$ <br\/>B\u01b0\u1edbc 2: T\u00ecm $m$ t\u1eeb gi\u1ea3 thi\u1ebft $B$ thu\u1ed9c parabol.<br\/>B\u01b0\u1edbc 3: T\u00ecm $n$ t\u1eeb gi\u1ea3 thi\u1ebft $C$ thu\u1ed9c parabol v\u1edbi $a$ v\u00e0 $m$ v\u1eeba t\u00ecm \u0111\u01b0\u1ee3c.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>\u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y=a{{x}^{2}}$\u0111i qua \u0111i\u1ec3m $A(3;3)$ n\u00ean ta c\u00f3 $3=a{{.3}^{2}}\\Leftrightarrow a=\\dfrac{1}{3}$ <br\/>V\u1eady h\u00e0m s\u1ed1 c\u00f3 d\u1ea1ng $y=\\dfrac{1}{3}{{x}^{2}}$ (P)<br\/>\u0110i\u1ec3m $B(-2;m)$ thu\u1ed9c (P) n\u00ean $m=\\dfrac{1}{3}.{{\\left( -2 \\right)}^{2}}\\Leftrightarrow m=\\dfrac{4}{3}$ <br\/>\u0110i\u1ec3m $C(n;1)$ thu\u1ed9c (P) n\u00ean $1=\\dfrac{1}{3}.{{n}^{2}}\\Leftrightarrow {{n}^{2}}=3\\Leftrightarrow n=\\pm \\sqrt{3}$<br\/>Do $n>m$ n\u00ean $n=\\sqrt{3}$<br\/>V\u1eady $m=\\dfrac{4}{3}$ v\u00e0 $n=\\sqrt{3}$.<\/span>"}]}],"id_ques":800},{"time":24,"part":[{"title":"\u0110i\u1ec1n c\u00e1c c\u1eb7p t\u1ecda \u0111\u1ed9 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng (\u0111\u1ec3 k\u1ebft qu\u1ea3 d\u1ea1ng s\u1ed1 th\u1eadp ph\u00e2n)","title_trans":"V\u00ed d\u1ee5: Ta t\u00ecm \u0111\u01b0\u1ee3c \u0111i\u1ec3m $A'(1;2)$ th\u00ec ta \u0111i\u1ec1n $(1;2).$<br\/><b>Ch\u00fa \u00fd:<\/b> \u0110i\u1ec1n c\u1ea3 d\u1ea5u ngo\u1eb7c v\u00e0 d\u1ea5u ;","temp":"fill_the_blank","correct":[[["(1;0,5)"],["(1;1)"],["(1;2)"]]],"list":[{"point":5,"width":70,"content":"","type_input":"","ques":"<span class='basic_left'>Cho ba h\u00e0m s\u1ed1 $y=\\dfrac{{{x}^{2}}}{2};y={{x}^{2}};y=2{{x}^{2}}$. Tr\u00ean m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 x\u00e1c \u0111\u1ecbnh ba \u0111i\u1ec3m $A\u2019;B\u2019;C\u2019$ \u0111\u1ed1i x\u1ee9ng v\u1edbi ba \u0111i\u1ec3m $A; B;C$ qua tr\u1ee5c $Oy$ c\u00f3 ho\u00e0nh \u0111\u1ed9 $x=-1$ theo th\u1ee9 t\u1ef1 n\u1eb1m tr\u00ean ba \u0111\u1ed3 th\u1ecb \u0111\u00e3 cho.<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> $A'$_input_; $B'$_input_;$C'$_input_","hint":"\u0110i\u1ec3m \u0111\u1ed1i x\u1ee9ng v\u1edbi $A(a;b)$ qua tr\u1ee5c $Oy$ c\u00f3 t\u1ecda \u0111\u1ed9 $(-a;b)$","explain":"<span class='basic_left'>B\u1ea3ng gi\u00e1 tr\u1ecb c\u1ee7a ba h\u00e0m s\u1ed1:<br\/><table> <tr> <td>$x$<\/td> <td>$-2$ <\/td> <td>$-1$<\/td><td>$0$<\/td><td>$1$<\/td> <td>$2$ <\/td><\/tr> <tr><td>$y=\\dfrac{{{x}^{2}}}{2}$<\/td><td>$2$ <\/td><td>$0,5$<\/td> <td>$0$<\/td><td>$0,5$<\/td><td>$2$<\/td><\/tr><tr><td>$y={{x}^{2}}$<\/td><td>$4$<\/td><td>$1$<\/td> <td>$0$<\/td><td>$1$<\/td><td>$4$<\/td><\/tr><tr><td>$y=2{{x}^{2}}$<\/td><td>$8$<\/td><td>$2$<\/td> <td>$0$<\/td><td>$2$<\/td><td>$8$<\/td><\/tr> <\/table><br\/>Ta c\u00f3 \u0111\u1ed3 th\u1ecb c\u1ee7a ba h\u00e0m s\u1ed1 tr\u00ean:<br\/><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai19/lv2/img\/TB6.5.png' \/><\/center><br\/><br\/>T\u1eeb \u0111\u1ed3 th\u1ecb c\u1ee7a ba h\u00e0m s\u1ed1, ta c\u00f3 $A(-1; 0,5),$ $B(-1;1 )$ v\u00e0 $C(-1; 2).$<br\/> Ba \u0111i\u1ec3m $A\u2019; B\u2019;C\u2019$ \u0111\u1ed1i x\u1ee9ng v\u1edbi $A; B;C$ qua tr\u1ee5c $Oy$ n\u00ean $A\u2019;B\u2019;C\u2019$ c\u0169ng l\u1ea7n l\u01b0\u1ee3t n\u1eb1m tr\u00ean ba \u0111\u1ed3 th\u1ecb $y=\\dfrac{{{x}^{2}}}{2};y={{x}^{2}};y=2{{x}^{2}}$<br\/>Suy ra $A(1; 0,5),$ $B(1;1 )$ v\u00e0 $C(1; 2).$<br\/><span class='basic_pink'>V\u1eady c\u00e1c c\u1eb7p t\u1ecda \u0111\u1ed9 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 $(1; 0,5),$ $(1;1 )$ v\u00e0 $(1; 2).$<\/span><br\/><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span> \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y=a{{x}^{2}}$ v\u1edbi $a>0$ th\u00ec v\u1edbi $a$ c\u00e0ng l\u1edbn, hai nh\u00e1nh c\u1ee7a parabol c\u00e0ng thu h\u1eb9p l\u1ea1i.<\/span><\/span>"}]}],"id_ques":801},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $n^2$","B. $n^3$","C. $n^4$"],"ques":"<span class='basic_left'>Cho h\u00e0m s\u1ed1 $y=-3{{x}^{2}}$ . N\u1ebfu $\\left| x \\right|$ g\u1ea5p $n$ l\u1ea7n th\u00ec $y$ g\u1ea5p ? l\u1ea7n<\/span>","hint":"","explain":"<span class='basic_left'>\u0110\u1ec3 bi\u1ec3u th\u1ecb $\\left| x \\right|$ g\u1ea5p $n$, ta cho $x$ hai gi\u00e1 tr\u1ecb $k$ v\u00e0 $nk$.<br\/>Khi \u0111\u00f3 c\u00e1c gi\u00e1 tr\u1ecb t\u01b0\u01a1ng \u1ee9ng c\u1ee7a $y$ l\u00e0 $-3{{k}^{2}}$ v\u00e0 $-3{{n}^{2}}{{k}^{2}}$<br\/> V\u1eady n\u1ebfu $\\left| x \\right|$ g\u1ea5p $n$ l\u1ea7n th\u00ec $y$ g\u1ea5p ${{n}^{2}}$ l\u1ea7n<\/span>"}]}],"id_ques":802},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<span class='basic_left'>Cho h\u00e0m s\u1ed1 $y=-2{{x}^{2}}$ c\u00f3 \u0111\u1ed3 th\u1ecb $(P).$ \u0110\u1ec3 \u0111\u01b0\u1eddng th\u1eb3ng $y=p$ c\u1eaft $(P)$ t\u1ea1i hai \u0111i\u1ec3m ph\u00e2n bi\u1ec7t th\u00ec $p$ th\u1ecfa m\u00e3n<\/span>","select":["A. $p>0$","B. $p<0$"],"hint":"D\u1ef1a v\u00e0o v\u1ecb tr\u00ed t\u01b0\u01a1ng \u0111\u1ed1i gi\u1eefa hai \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y=-2{{x}^{2}}$ v\u00e0 $y=p$ \u0111\u1ec3 t\u00ecm $p$ ","explain":"<span class='basic_left'>Ta c\u00f3 b\u1ea3ng gi\u00e1 tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1:<br\/><table> <tr> <td>$x$<\/td> <td>$-2$ <\/td> <td>$-1$<\/td><td>$0$<\/td><td>$1$<\/td> <td>$2$ <\/td><\/tr> <tr><td>$y=-2{{x}^{2}}$<\/td><td>$-8$<\/td><td>$-2$<\/td> <td>$0$<\/td><td>$-2$<\/td><td>$-8$<\/td><\/tr> <\/table><br\/>\u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y=-2{{x}^{2}}$ v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng $y=p$:<br\/><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai19/lv2/img\/TB6.7.png' \/><\/center><br\/><br\/>D\u1ef1a v\u00e0o \u0111\u1ed3 th\u1ecb ta c\u00f3: \u0110\u1ec3 \u0111\u01b0\u1eddng th\u1eb3ng $y=p$ c\u1eaft $(P)$ t\u1ea1i hai \u0111i\u1ec3m ph\u00e2n bi\u1ec7t th\u00ec $p<0$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 B<\/span><\/span>","column":2}]}],"id_ques":803},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<span class='basic_left'>Cho h\u00e0m s\u1ed1 $y=\\left( \\sqrt{m+1}-2 \\right){{x}^{2}}$ .<br\/> Khi $x>0,$ h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn n\u1ebfu<\/span> ","select":["A. $m>3$","B. $m>3$ ho\u1eb7c $m<-1$ ","C. $m<3$","D. $-1\\le m<3$ "],"hint":"","explain":"<span class='basic_left'>H\u00e0m s\u1ed1 $y=\\left( \\sqrt{m+1}-2 \\right){{x}^{2}}$c\u00f3 d\u1ea1ng $y=a{{x}^{2}}$ <br\/>Khi $x>0,$ h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn n\u1ebfu $a>0\\Leftrightarrow \\sqrt{m+1}-2>0\\Leftrightarrow \\sqrt{m+1}>2$ <br\/>$\\Leftrightarrow \\left\\{ \\begin{aligned} & m+1\\ge 0 \\\\ & m+1>4 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m\\ge -1 \\\\ & m>3 \\\\ \\end{aligned} \\right.\\Leftrightarrow m>3$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A<\/span><\/span>","column":2}]}],"id_ques":804},{"time":24,"part":[{"title":"\u0110i\u1ec1n c\u00e1c c\u1eb7p t\u1ecda \u0111\u1ed9 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng ","title_trans":"V\u00ed d\u1ee5: Ta t\u00ecm \u0111\u01b0\u1ee3c \u0111i\u1ec3m $M(1;2)$ th\u00ec ta \u0111i\u1ec1n $(1;2).$<br\/><b>Ch\u00fa \u00fd:<\/b> \u0110i\u1ec1n c\u1ea3 d\u1ea5u ngo\u1eb7c v\u00e0 d\u1ea5u ;","temp":"fill_the_blank_random","correct":[[["(0;0)"],["(4;-8)"],["(-4;-8)"]]],"list":[{"point":5,"width":70,"content":"","type_input":"","ques":"<span class='basic_left'>Cho h\u00e0m s\u1ed1 $y=-\\dfrac{1}{2}{{x}^{2}}$ (P)<br\/>T\u00ecm t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m $M$ thu\u1ed9c parabol n\u00f3i tr\u00ean bi\u1ebft r\u1eb1ng kho\u1ea3ng c\u00e1ch t\u1eeb $M$ \u0111\u1ebfn tr\u1ee5c ho\u00e0nh g\u1ea5p \u0111\u00f4i kho\u1ea3ng c\u00e1ch t\u1eeb $M$ \u0111\u1ebfn tr\u1ee5c tung.<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> C\u00e1c \u0111i\u1ec3m th\u1ecfa m\u00e3n \u0111\u1ec1 b\u00e0i l\u00e0 _input_; _input_; _input_","hint":"T\u00ecm xem qu\u1ef9 t\u00edch c\u00e1c \u0111i\u1ec3m M l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng n\u00e0o? ","explain":"<span class='basic_left'>V\u00ec kho\u1ea3ng c\u00e1ch t\u1eeb $M$ \u0111\u1ebfn tr\u1ee5c $Ox$ g\u1ea5p \u0111\u00f4i kho\u1ea3ng c\u00e1ch t\u1eeb $M$ \u0111\u1ebfn tr\u1ee5c $Oy$ n\u00ean tung \u0111\u1ed9 c\u1ee7a $M$ b\u1eb1ng $\\pm 2$ l\u1ea7n ho\u00e0nh \u0111\u1ed9 c\u1ee7a $M$.<br\/> Do \u0111\u00f3 t\u1eadp h\u1ee3p \u0111i\u1ec3m $M$ l\u00e0 hai \u0111\u01b0\u1eddng th\u1eb3ng $y=2x$ v\u00e0 $ y=-2x$<br\/>Do $M$ thu\u1ed9c $(P)$ n\u00ean t\u1ecda \u0111\u1ed9 $M$ l\u00e0 nghi\u1ec7m c\u1ee7a hai h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh:<\/span><br\/>$\\left\\{ \\begin{aligned} & y=-\\dfrac{1}{2}{{x}^{2}} \\\\ & y=2x \\\\ \\end{aligned} \\right.$ (I) v\u00e0 $\\left\\{ \\begin{aligned} & y=-\\dfrac{1}{2}{{x}^{2}} \\\\ & y=-2x \\\\ \\end{aligned} \\right.$(II)<br\/><span class='basic_left'>Gi\u1ea3i h\u1ec7 (I)<br\/>(I) $\\Leftrightarrow \\left\\{ \\begin{aligned} & -\\dfrac{1}{2}{{x}^{2}}=2x \\\\ & y=2x \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & {{x}^{2}}+4x=0 \\\\ & y=2x \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x\\left( x+4 \\right)=0 \\\\ & y=2x \\\\ \\end{aligned} \\right.$ <br\/>$\\Leftrightarrow \\left\\{ \\begin{aligned} & \\left[ \\begin{aligned} & x=0 \\\\ & x=-4 \\\\ \\end{aligned} \\right. \\\\ & y=2x \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left[ \\begin{aligned} & x=y=0 \\\\ & x=-4;y=-8 \\\\ \\end{aligned} \\right.$<br\/>T\u01b0\u01a1ng t\u1ef1 gi\u1ea3i h\u1ec7 (II), ta c\u00f3 $x=y=0$ v\u00e0 $x=4$;$y=-8$<br\/>V\u1eady c\u00f3 ba \u0111i\u1ec3m th\u1ecfa m\u00e3n \u0111\u1ec1 b\u00e0i l\u00e0 $(0;0); (4;-8)$ v\u00e0 $(-4;-8)$<br\/><span class='basic_pink'>V\u1eady c\u00e1c c\u1eb7p t\u1ecda \u0111\u1ed9 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $(0;0); (4;-8)$ v\u00e0 $(-4;-8)$<\/span><\/span>"}]}],"id_ques":805},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Cho h\u00e0m s\u1ed1 $y=\\left( 2m-1 \\right){{x}^{2}}$ c\u00f3 \u0111\u1ed3 th\u1ecb \u0111i qua $(3;-3).$ M\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi tr\u1ee5c ho\u00e0nh c\u1eaft tr\u1ee5c tung t\u1ea1i \u0111i\u1ec3m $-4$ v\u00e0 c\u1eaft parabol n\u00f3i tr\u00ean t\u1ea1i c\u00e1c \u0111i\u1ec3m $A, B.$<br\/> Di\u1ec7n t\u00edch tam gi\u00e1c $OAB$ l\u00e0","select":["A. $2\\sqrt{3}$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)","B. $8\\sqrt{3}$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)","C. $5\\sqrt{3}$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)","D. $6\\sqrt{3}$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)"],"hint":"T\u00ecm t\u1ecda \u0111\u1ed9 hai \u0111i\u1ec3m $A, B$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: T\u00ecm $m$ t\u1eeb gi\u1ea3 thi\u1ebft: H\u00e0m s\u1ed1 c\u00f3 \u0111\u1ed3 th\u1ecb \u0111i qua $(3;-3).$<br\/>B\u01b0\u1edbc 2: V\u1ebd \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 t\u00ecm \u0111\u01b0\u1ee3c, t\u1eeb \u0111\u00f3 x\u00e1c \u0111\u1ecbnh $A, B$<br\/>B\u01b0\u1edbc 3: T\u00ednh $AB$ v\u00e0 t\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c $OAB$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y=\\left( 2m-1 \\right){{x}^{2}}$\u0111i qua $(3;-3)$ n\u00ean ta c\u00f3:<br\/>$-3=\\left( 2m-1 \\right){{.3}^{2}}\\Leftrightarrow 2m-1=-\\dfrac{1}{3}\\Leftrightarrow m=\\dfrac{1}{3}$ <br\/>Suy ra h\u00e0m s\u1ed1 t\u00ecm \u0111\u01b0\u1ee3c l\u00e0 $y=-\\dfrac{1}{3}{{x}^{2}}$<br\/>\u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y=-\\dfrac{1}{3}{{x}^{2}}$ <br\/><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai19/lv2/img\/TB6.4.png' \/><\/center> <br\/><br\/>Ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m A v\u00e0 B l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh <br\/>$\\dfrac{-1}{3}{{x}^{2}}=-4\\Leftrightarrow x=\\pm 2\\sqrt{3}$<br\/>Gi\u1ea3 s\u1eed $A\\left( -2\\sqrt{3};-4 \\right)$ v\u00e0 $B\\left( 2\\sqrt{3};-4 \\right)$ nh\u01b0 h\u00ecnh v\u1ebd<br\/> Ta c\u00f3 $AB=2\\sqrt{3}-\\left( -2\\sqrt{3} \\right)=4\\sqrt{3}$<br\/>Suy ra ${{S}_{OAB}}=\\dfrac{1}{2}.4\\sqrt{3}.\\left| -4 \\right|=8\\sqrt{3}$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch) <br\/> V\u1eady di\u1ec7n t\u00edch tam gi\u00e1c $OAB$ l\u00e0 $8\\sqrt{3}$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span> <\/span>","column":2}]}],"id_ques":806},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-64"],["0"]]],"list":[{"point":5,"width":50,"content":"","type_input":"","ques":"Cho h\u00e0m s\u1ed1 $y=a{{x}^{2}}$ . Bi\u1ebft r\u1eb1ng khi $x=5$ th\u00ec $y=-100.$ T\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t v\u00e0 l\u1edbn nh\u1ea5t c\u1ee7a $y$ khi $x$ th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n $-2\\le x\\le 4$ <br\/><b> \u0110\u00e1p s\u1ed1: <\/b> Gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a $y$ l\u00e0 _input_ v\u00e0 gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a $y$ l\u00e0 _input_","hint":"X\u00e9t t\u00ednh \u0111\u1ed3ng bi\u1ebfn ngh\u1ecbch bi\u1ebfn c\u1ee7a h\u00e0m s\u1ed1 khi $-2\\le x\\le 0$ v\u00e0 $0\\le x\\le 4$ ","explain":"<span class='basic_left'> Khi $x=5$ th\u00ec $y=-100$ n\u00ean $-100=a{{.5}^{2}}\\Leftrightarrow a=-4$ <br\/>V\u1eady $y=-4{{x}^{2}}$ <br\/> H\u00e0m s\u1ed1 $y=-4{{x}^{2}}$c\u00f3 $a=-4<0$ n\u00ean $y=0$ l\u00e0 gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1 v\u00e0 h\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn khi $x>0,$ \u0111\u1ed3ng bi\u1ebfn khi $x<0.$<br\/>X\u00e9t $-2\\le x\\le 0$, h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn n\u00ean<br\/> $f\\left( -2 \\right)\\le f\\left( x \\right)\\le f\\left( 0 \\right)$ hay $-16\\le f\\left( x \\right)\\le 0$ (1)<br\/>X\u00e9t $0\\le x\\le 4$, h\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn n\u00ean<br\/> $f\\left( 0 \\right)\\ge f\\left( x \\right)\\ge f\\left( 4 \\right)$ hay $-64\\le f\\left( x \\right)\\le 0$ (2)<br\/>Suy ra khi $-2\\le x\\le 4$ th\u00ec gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1 l\u00e0 $0$ v\u00e0 gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1 l\u00e0 $-64.$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0 $-64$ v\u00e0 $0$<\/span><\/span>"}]}],"id_ques":807},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" N\u1ebfu \u0111i\u1ec3m $A(c;d)$ thu\u1ed9c \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 $y=4{{x}^{2}}$ th\u00ec \u0111i\u1ec3m $B(2c;d)$ thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y={{x}^{2}}$, <b>\u0111\u00fang<\/b> hay <b> sai <\/b>?","select":["A. \u0110\u00fang","B. Sai"],"hint":"","explain":"<span class='basic_left'>N\u1ebfu \u0111i\u1ec3m $A(c;d)$ thu\u1ed9c \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 $y=4{{x}^{2}}$ th\u00ec $d=4{{c}^{2}}$ <br\/> Suy ra $d={{\\left( 2c \\right)}^{2}}$ <br\/>Suy ra \u0111i\u1ec3m $B(2c;d)$ c\u00f3 t\u1ecda \u0111\u1ed9 th\u1ecfa m\u00e3n \u0111\u1eb3ng th\u1ee9c $y={{x}^{2}}$ <br\/>Do \u0111\u00f3 \u0111i\u1ec3m $B(2c;d)$ thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y={{x}^{2}}$.<br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 \u0111\u00fang.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A<\/span><\/span>","column":2}]}],"id_ques":808},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u $(>,=,<)$ th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[[">"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"<span class='basic_left'>Cho h\u00e0m s\u1ed1 $y=f\\left( x \\right)=a{{x}^{2}}$ . Bi\u1ebft $x=\\dfrac{1}{4}$ th\u00ec $y=\\dfrac{3}{16}$ <br\/> V\u1edbi ${{x}_{1}};$ ${{x}_{2}}$ l\u00e0 nh\u1eefng s\u1ed1 \u00e2m v\u00e0 ${{x}_{1}}<{{x}_{2}}$ th\u00ec $f\\left( {{x}_{1}} \\right)$ _input_$f\\left( {{x}_{2}} \\right)$ <\/span>","hint":"T\u00ecm $a$ r\u1ed3i x\u00e9t t\u00ednh \u0111\u1ed3ng bi\u1ebfn v\u00e0 ngh\u1ecbch bi\u1ebfn c\u1ee7a h\u00e0m s\u1ed1 khi $x<0$ ","explain":"<span class='basic_left'>Ta c\u00f3 $x=\\dfrac{1}{4}$ th\u00ec $y=\\dfrac{3}{16}$ n\u00ean<br\/> $\\dfrac{3}{16}=a.{{\\left( \\dfrac{1}{4} \\right)}^{2}}\\Leftrightarrow a=3$ <br\/>V\u1eady h\u00e0m s\u1ed1 t\u00ecm \u0111\u01b0\u1ee3c l\u00e0 $y=3{{x}^{2}}$ <br\/>Do $a=3>0$ n\u00ean h\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn khi $x<0$<br\/>Suy ra v\u1edbi ${{x}_{1}}<{{x}_{2}}<0$ th\u00ec $f\\left( {{x}_{1}} \\right)>f\\left( {{x}_{2}} \\right)$ <br\/><span class='basic_pink'>V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $>$<\/span><\/span>"}]}],"id_ques":809},{"time":24,"part":[{"title":"Cho h\u00e0m s\u1ed1 $y=a{{x}^{2}}\\left( a\\ne 0 \\right),$ \u0111i\u1ec3m $C\\left( m;n \\right)$ thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 \u0111\u00f3 $\\left( m\\ne n \\right).$ ","title_trans":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","temp":"true_false","correct":[["t","t","f","f","t"]],"list":[{"point":5,"image":"img\/1.png","col_name":["","\u0110\u00fang","Sai"],"arr_ques":["a. \u0110i\u1ec3m \u0111\u1ed1i x\u1ee9ng v\u1edbi $C$ qua tr\u1ee5c $Ox$ kh\u00f4ng thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1.","b. \u0110i\u1ec3m \u0111\u1ed1i x\u1ee9ng v\u1edbi $C$ qua tr\u1ee5c $Oy$ thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1.","c. \u0110i\u1ec3m $\\left( m;-n \\right)$ thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1.","d. \u0110i\u1ec3m $\\left( -m;-n \\right)$ thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1","e. \u0110i\u1ec3m $\\left( -m;n \\right)$ thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1"],"hint":"D\u1ef1a v\u00e0o t\u00ednh ch\u1ea5t: \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y=a{{x}^{2}}\\left( a\\ne 0 \\right)$ \u0111i qua $O$ v\u00e0 nh\u1eadn $Oy$ l\u00e0m tr\u1ee5c \u0111\u1ed1i x\u1ee9ng.","explain":["a. \u0110\u00fang v\u00ec \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y=a{{x}^{2}}\\left( a\\ne 0 \\right)$ \u0111i qua $O$ v\u00e0 nh\u1eadn $Oy$ l\u00e0m tr\u1ee5c \u0111\u1ed1i x\u1ee9ng n\u00ean \u0111i\u1ec3m \u0111\u1ed1i x\u1ee9ng v\u1edbi $C$ qua tr\u1ee5c $Ox$ kh\u00f4ng thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1.","<br\/>b. \u0110\u00fang v\u00ec \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y=a{{x}^{2}}\\left( a\\ne 0 \\right)$ \u0111i qua $O$ v\u00e0 nh\u1eadn $Oy$ l\u00e0m tr\u1ee5c \u0111\u1ed1i x\u1ee9ng n\u00ean \u0111i\u1ec3m \u0111\u1ed1i x\u1ee9ng v\u1edbi $C$ qua tr\u1ee5c $Oy$ thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1.","<br\/>c. Sai v\u00ec \u0111i\u1ec3m$\\left( m;-n \\right)$ \u0111\u1ed1i x\u1ee9ng v\u1edbi $C$ qua tr\u1ee5c $Ox$ n\u00ean kh\u00f4ng thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1","<br\/>d. Sai v\u00ec \u0111i\u1ec3m $(-m;-n)$ \u0111\u1ed1i x\u1ee9ng v\u1edbi $C$ qua g\u1ed1c t\u1ecda \u0111\u1ed9 $O$ n\u00ean kh\u00f4ng thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1","<br\/> e. \u0110\u00fang v\u00ec \u0111i\u1ec3m $\\left( -m;n \\right)$ \u0111\u1ed1i x\u1ee9ng v\u1edbi $C$ qua $Oy$ n\u00ean thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1"]}]}],"id_ques":810}],"lesson":{"save":0,"level":2}}