{"segment":[{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. ($\\dfrac{7}{3};-\\dfrac{2}{3}$)","B. ($\\dfrac{4}{5};\\dfrac{2}{3}$)","C. ($\\dfrac{1}{3};\\dfrac{7}{3}$)"],"ques":"Gi\u1ea3i h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh $\\left\\{ \\begin{align} & 5\\left( x-y \\right)-3\\left( x+2y \\right)=12 \\\\ & 3\\left( x-y \\right)-4\\left( x+2y \\right)=5 \\\\ \\end{align} \\right.$ <br\/><b> \u0110\u00e1p s\u1ed1: <\/b> $(x;y)$ = (?;?) ","explain":"<span class='basic_left'>R\u00fat g\u1ecdn t\u1eebng ph\u01b0\u01a1ng tr\u00ecnh ta c\u00f3 h\u1ec7: <br\/> $\\left\\{ \\begin{align} & 5\\left( x-y \\right)-3\\left( x+2y \\right)=12 \\\\ & 3\\left( x-y \\right)-4\\left( x+2y \\right)=5 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & 2x-11y=12 \\\\ & -x-11y=5 \\\\ \\end{align} \\right.$ <br\/> C\u1ed9ng t\u1eebng v\u1ebf c\u1ee7a hai ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c: <br\/> $\\Leftrightarrow \\left\\{ \\begin{align} & 3x=7 \\\\ & -x-11y=5 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & x=\\dfrac{7}{3} \\\\ & -\\dfrac{7}{3}-11y=5 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & x=\\dfrac{7}{3} \\\\ & y=-\\dfrac{2}{3} \\\\ \\end{align} \\right.$ <br\/> Suy ra $\\left(\\dfrac{7}{3};-\\dfrac{2}{3}\\right)$ l\u00e0 nghi\u1ec7m c\u1ee7a h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh <br\/><br\/> <b> L\u01b0u \u00fd: <\/b> Ngo\u00e0i c\u00e1ch gi\u1ea3i tr\u1ef1c ti\u1ebfp, h\u1ecdc sinh c\u00f3 th\u1ec3 v\u1eadn d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p \u0111\u1eb7t \u1ea9n ph\u1ee5 \u0111\u1ec3 gi\u1ea3i b\u00e0i to\u00e1n tr\u00ean."}]}],"id_ques":291},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","temp":"fill_the_blank","correct":[[["36"],["12"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" T\u00ecm nghi\u1ec7m c\u1ee7a h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh $\\left\\{ \\begin{align} & \\dfrac{3}{5x}+\\dfrac{1}{y}=\\dfrac{1}{10} \\\\ & \\dfrac{3}{4x}+\\dfrac{3}{4y}=\\dfrac{1}{12} \\\\ \\end{align} \\right.$ <\/b> <br\/><b> \u0110\u00e1p s\u1ed1: <\/b> $(x;y) = $ (_input_;_input_) ","hint":"D\u00f9ng ph\u01b0\u01a1ng ph\u00e1p \u0111\u1eb7t \u1ea9n ph\u1ee5 r\u1ed3i \u00e1p d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p th\u1ebf ho\u1eb7c c\u1ed9ng \u0111\u1ea1i s\u1ed1 \u0111\u1ec3 gi\u1ea3i h\u1ec7","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $\\left\\{ \\begin{align} & x\\ne 0 \\\\ & y\\ne 0 \\\\ \\end{align} \\right.$ <br\/> \u0110\u1eb7t $\\left\\{ \\begin{align} & \\dfrac{1}{x}=u \\\\ & \\dfrac{1}{y}=v \\\\ \\end{align} \\right.$ <br\/> H\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh: $\\left\\{ \\begin{align} & \\dfrac{3u}{5}+v=\\dfrac{1}{10} \\\\ & \\dfrac{3}{4}u+\\dfrac{3}{4}v=\\dfrac{1}{12} \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & \\dfrac{9u}{20}+\\dfrac{3}{4}v=\\dfrac{3}{40}\\,(1) \\\\ & \\dfrac{3}{4}u+\\dfrac{3}{4}v=\\dfrac{1}{12}\\,(2) \\\\ \\end{align} \\right.$ <br\/> L\u1ea5y (2) $-$ (1) $\\Leftrightarrow \\left\\{ \\begin{align} & \\dfrac{3u}{10}=\\dfrac{1}{120} \\\\ & v=\\dfrac{1}{10}-\\dfrac{3u}{5} \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & u=\\dfrac{1}{36} \\\\ & v=\\dfrac{1}{10}-\\dfrac{3}{5}.\\dfrac{1}{36} \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & u=\\dfrac{1}{36} \\\\ & v=\\dfrac{1}{12} \\\\ \\end{align} \\right.$ <br\/> $\\Leftrightarrow \\left\\{ \\begin{align} & \\dfrac{1}{x}=\\dfrac{1}{36} \\\\ & \\dfrac{1}{y}=\\dfrac{1}{12} \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & x=36 \\\\ & y=12 \\\\ \\end{align} \\right.$ <br\/> Suy ra $(36;12)$ l\u00e0 nghi\u1ec7m c\u1ee7a h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh <br\/> <span class='basic_pink'> V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u1ea7n l\u01b0\u1ee3t l\u00e0 $36$ v\u00e0 $12$ <\/span> "}]}],"id_ques":292},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"select":["A. ($\\dfrac{11}{40};\\dfrac{2}{3}$)","B. ($\\dfrac{11}{40};-\\dfrac{9}{20}$)","C. ($\\dfrac{11}{40};\\dfrac{7}{3}$)"],"ques":" Gi\u1ea3i h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh sau: <br\/> $\\left\\{ \\begin{align} & \\dfrac{7}{2x+y}+\\dfrac{4}{2x-y}=74 \\\\ & \\dfrac{3}{2x+y}+\\dfrac{2}{2x-y}=32 \\\\ \\end{align} \\right.$ <br\/><b> \u0110\u00e1p s\u1ed1: <\/b> $(x;y) = $ (?;?) ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $2x\\ne \\pm y$ <br\/> \u0110\u1eb7t $\\left\\{ \\begin{align} & \\dfrac{1}{2x+y}=u \\\\ & \\dfrac{1}{2x-y}=v \\\\ \\end{align} \\right.$ <br\/> H\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh: $\\left\\{ \\begin{align} & 7u+4v=74 \\\\ & 3u+2v=32 \\\\ \\end{align} \\right.$ <br\/> $\\Leftrightarrow \\left\\{ \\begin{align} & 7u+4v=74 \\\\ & 6u+4v=64 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & u=10 \\\\ & 3.10+2v=32 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & u=10 \\\\ & 2v=2 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & u=10 \\\\ & v=1 \\\\ \\end{align} \\right.$ <br\/> V\u1edbi $ \\left\\{ \\begin{align} & u=10 \\\\ & v=1 \\\\ \\end{align} \\right.$ $\\Leftrightarrow\\left\\{ \\begin{align} & \\dfrac{1}{2x+y}=10 \\\\ & \\dfrac{1}{2x-y}=1 \\\\ \\end{align} \\right.$ <br\/> $\\Leftrightarrow \\left\\{ \\begin{align} & 2x+y=\\dfrac{1}{10} \\\\ & 2x-y=1 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & 4x=\\dfrac{11}{10} \\\\ & 2x-y=1 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & x=\\dfrac{11}{40} \\\\ & 2.\\dfrac{11}{40}-y=1 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & x=\\dfrac{11}{40} \\\\ & y=\\dfrac{-9}{20} \\\\ \\end{align} \\right.$ <br\/> Suy ra $\\left(\\dfrac{11}{40};\\dfrac{-9}{20}\\right)$ l\u00e0 nghi\u1ec7m c\u1ee7a h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh <br\/> "}]}],"id_ques":293},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"select":["A. ($\\dfrac{1}{7};\\dfrac{1}{9};\\dfrac{1}{3}$)","B. ($\\dfrac{1}{8};\\dfrac{1}{7};\\dfrac{1}{9}$)","C. ($\\dfrac{1}{7};\\dfrac{1}{9};\\dfrac{1}{11}$)"],"ques":"Gi\u1ea3i h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh $\\left\\{ \\begin{align} & \\dfrac{1}{x}+\\dfrac{1}{y}=16 \\\\ & \\dfrac{1}{y}+\\dfrac{1}{z}=20 \\\\ & \\dfrac{1}{z}+\\dfrac{1}{x}=18 \\\\ \\end{align} \\right.$ <br\/><b> \u0110\u00e1p s\u1ed1: <\/b> $(x;y;z)$ = (?;?;?) ","hint":"C\u1ed9ng t\u1eebng v\u1ebf c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a h\u1ec7 v\u1edbi nhau.","explain":"<span class='basic_left'>X\u00e9t h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh $\\left\\{ \\begin{align} & \\dfrac{1}{x}+\\dfrac{1}{y}=16\\,(1) \\\\ & \\dfrac{1}{y}+\\dfrac{1}{z}=20\\,(2) \\\\ & \\dfrac{1}{z}+\\dfrac{1}{x}=18\\,(3) \\\\ \\end{align} \\right.$ <br\/> \u0110i\u1ec1u ki\u1ec7n: $x\\ne 0; y\\ne 0; z\\ne 0$ <br\/> C\u1ed9ng t\u1eebng v\u1ebf c\u1ee7a c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh ta \u0111\u01b0\u1ee3c: <br\/> $ 2.\\left( \\dfrac{1}{x}+\\dfrac{1}{y}+\\dfrac{1}{z} \\right)=54\\Leftrightarrow \\dfrac{1}{x}+\\dfrac{1}{y}+\\dfrac{1}{z}=27\\,\\left( * \\right)$ <br\/> Th\u1ebf ph\u01b0\u01a1ng tr\u00ecnh (2) v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh (*), ta \u0111\u01b0\u1ee3c: <br\/> $ \\dfrac{1}{x}+20=27\\Leftrightarrow \\dfrac{1}{x}=7(**)$ <br\/> Th\u1ebf ph\u01b0\u01a1ng tr\u00ecnh (**) v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh (1) v\u00e0 (3), ta \u0111\u01b0\u1ee3c: <br\/> $\\left\\{ \\begin{align} & \\dfrac{1}{z}+7=18 \\\\ & 7+\\dfrac{1}{y}=16 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & \\dfrac{1}{z}=11 \\\\ & \\dfrac{1}{y}=9 \\\\ \\end{align} \\right.$$\\left( *** \\right)$ <br\/> T\u1eeb (**) v\u00e0 (***), suy ra $\\left\\{ \\begin{align}& x=\\dfrac{1}{7} \\\\ & y=\\dfrac{1}{9} \\\\ & z=\\dfrac{1}{11} \\\\ \\end{align} \\right.$<br\/>"}]}],"id_ques":294},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","temp":"fill_the_blank","correct":[[["-2"],["3"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" T\u00ecm gi\u00e1 tr\u1ecb c\u1ee7a $m$ v\u00e0 $n$ \u0111\u1ec3 \u0111a th\u1ee9c sau b\u1eb1ng $0$ v\u1edbi m\u1ecdi $x$ <br\/> $P\\left( x \\right)=\\left( 2m+3n-5 \\right)x+5m+3n+1$ <br\/><b> \u0110\u00e1p s\u1ed1: <\/b> $(m;n) = $ (_input_;_input_) ","hint":"\u0110a th\u1ee9c m\u1ed9t \u1ea9n b\u1eb1ng $0$ khi v\u00e0 ch\u1ec9 khi c\u00e1c h\u1ec7 s\u1ed1 c\u1ee7a \u0111a th\u1ee9c \u0111\u1ec1u \u0111\u1ed3ng nh\u1ea5t b\u1eb1ng $0$","explain":"<span class='basic_left'> Ta c\u00f3: <br\/> $P\\left( x \\right)=0\\,\\,\\forall x $ <br\/> $ \\Leftrightarrow \\left( 2m+3n-5 \\right)x+5m+3n+1=0\\,\\,\\forall x $ <br\/> $\\Leftrightarrow \\left\\{ \\begin{align} & 2m+3n-5=0 \\\\ & 5m+3n+1=0 \\\\ \\end{align} \\right.$ $\\Leftrightarrow \\left\\{ \\begin{align} & 2m+3n=5 \\\\ & 5m+3n=-1 \\\\ \\end{align} \\right.$ <br\/> L\u1ea5y t\u1eebng v\u1ebf ph\u01b0\u01a1ng tr\u00ecnh th\u1ee9 nh\u1ea5t c\u1ee7a h\u1ec7 tr\u1eeb \u0111i t\u1eebng v\u1ebf ph\u01b0\u01a1ng tr\u00ecnh th\u1ee9 hai, ta \u0111\u01b0\u1ee3c: <br\/> $ \\left\\{ \\begin{align} & -3m=6 \\\\ & 2m+3n=5 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & m=-2 \\\\ & 2.\\left( -2 \\right)+3n=5 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & m=-2 \\\\ & n=3 \\\\ \\end{align} \\right.$ <br\/> <span class='basic_pink'> V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u1ea7n l\u01b0\u1ee3t l\u00e0 $-2$ v\u00e0 $3$ <\/span> "}]}],"id_ques":295},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t"," title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"Cho h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh $\\left\\{ \\begin{align} & x + y =1 \\\\ & x^3 + y^3 = x^2+y^2 \\\\ \\end{align} \\right.$ <br\/> Nghi\u1ec7m c\u1ee7a h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $(0;1)$ ","B. $(1;0)$","C. $(0;1)$ v\u00e0 $(1;0)$","D. $(0;1)$; $(1;0)$ v\u00e0 $(1;1)$"],"explain":"<span class='basic_left'> Ta c\u00f3: $\\left\\{ \\begin{align} & x+y=1 \\\\ & {{x}^{3}}+{{y}^{3}}={{x}^{2}}+{{y}^{2}} \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & x+y=1\\,\\left( 1 \\right) \\\\ & \\left( x+y \\right)\\left( {{x}^{2}}+xy+{{y}^{2}}\\right)={{x}^{2}}+{{y}^{2}}\\,\\left( 2 \\right) \\\\ \\end{align} \\right.$ <br\/> Th\u1ebf ph\u01b0\u01a1ng tr\u00ecnh (1) v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh (2), ta \u0111\u01b0\u1ee3c: <br\/> $\\left\\{ \\begin{align} & x+y=1 \\\\ & {{x}^{2}}+xy+{{y}^{2}}={{x}^{2}}+{{y}^{2}} \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & x+y=1 \\\\ & xy=0 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{aligned} & x+y=1 \\\\ & \\left[ \\begin{aligned} & x=0 \\\\ & y=0 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned} \\right.$<br\/> $\\bullet$ V\u1edbi $x = 0 \\Rightarrow 0 + y = 1 \\Rightarrow y = 1$ <br\/> $\\bullet$ V\u1edbi $y = 0 \\Rightarrow x + 0 = 1 \\Rightarrow x = 1$ <br\/> Suy ra $(0;1)$ v\u00e0 $(1;0)$ l\u00e0 c\u00e1c nghi\u1ec7m c\u1ee7a h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C <\/span> <br\/> <b> Nh\u1eadn x\u00e9t<\/b> <br\/> H\u1ec7 \u0111\u1ed1i x\u1ee9ng lo\u1ea1i I l\u00e0 h\u1ec7 ch\u1ee9a 2 \u1ea9n $x,y$ m\u00e0 khi ta thay \u0111\u1ed5i vai tr\u00f2 $x,y$ cho nhau th\u00ec h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh kh\u00f4ng thay \u0111\u1ed5i. <br\/> \u0110\u1ed1i v\u1edbi lo\u1ea1i ph\u01b0\u01a1ng tr\u00ecnh n\u00e0y, ta th\u01b0\u1eddng \u00e1p d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p th\u1ebf \u0111\u1ec3 gi\u1ea3i. ","column":2}]}],"id_ques":296},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t"," title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"D\u00f9ng h\u00ecnh v\u1ebd x\u00e1c \u0111\u1ecbnh s\u1ed1 nghi\u1ec7m c\u1ee7a h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh sau: $\\left\\{ \\begin{align} & 1,5y+x=-0,5 \\\\ & x+2y=4 \\\\ & -2x-4y=10 \\\\ \\end{align} \\right.$","select":["A. V\u00f4 nghi\u1ec7m ","B. V\u00f4 s\u1ed1 nghi\u1ec7m","C. M\u1ed9t nghi\u1ec7m duy nh\u1ea5t"],"hint":"N\u1ebfu ba \u0111\u01b0\u1eddng th\u1eb3ng bi\u1ec3u di\u1ec5n t\u1eadp nghi\u1ec7m c\u1ee7a c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh trong h\u1ec7 c\u00f9ng \u0111i qua m\u1ed9t \u0111i\u1ec3m th\u00ec \u0111i\u1ec3m \u0111\u00f3 l\u00e0 nghi\u1ec7m c\u1ee7a h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh","explain":"<span class='basic_left'> Bi\u1ebfn \u0111\u1ed5i h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh ta c\u00f3: <br\/> $\\left\\{ \\begin{align} & 1,5y+x=-0,5 \\\\ & x+2y=4 \\\\ & -2x-4y=10 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & y=\\dfrac{-2}{3}x-\\dfrac{1}{3}\\,\\left( {{d}_{1}} \\right) \\\\ & y=\\dfrac{-1}{2}x+2\\,\\left( {{d}_{2}} \\right) \\\\ & y=\\dfrac{-1}{2}x-\\dfrac{5}{2}\\,\\left( {{d}_{3}} \\right) \\\\ \\end{align} \\right.$ <br\/> Bi\u1ec3u di\u1ec5n \u0111\u1ed3 th\u1ecb c\u1ee7a c\u00e1c \u0111\u01b0\u1eddng th\u1eb3ng $(d_1); (d_2)$ v\u00e0 $(d_3)$ ta c\u00f3 h\u00ecnh v\u1ebd: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai15/lv3/img\/h932_k1.png' \/><\/center> <br\/> V\u00ec $(d_2) \/\/ (d_3)$ n\u00ean hai \u0111\u01b0\u1eddng th\u1eb3ng kh\u00f4ng c\u00f3 \u0111i\u1ec3m chung <br\/> $\\Rightarrow$ Ba \u0111\u01b0\u1eddng th\u1eb3ng kh\u00f4ng c\u00f9ng \u0111i qua m\u1ed9t \u0111i\u1ec3m chung n\u00e0o <br\/> $\\Rightarrow$ H\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m. <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A <\/span>","column":3}]}],"id_ques":297},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","temp":"fill_the_blank","correct":[[["3"],["-6"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" <span class='basic_left'>Bi\u1ebft r\u1eb1ng: <b> \u0110a th\u1ee9c $P(x)$ chia h\u1ebft cho \u0111a th\u1ee9c $x - a$ khi v\u00e0 ch\u1ec9 khi $P(a) = 0.$ <\/b> H\u00e3y t\u00ecm c\u00e1c gi\u00e1 tr\u1ecb c\u1ee7a $m$ v\u00e0 $n$ sao cho \u0111a th\u1ee9c sau chia h\u1ebft cho $x^2+2x:$ <br\/> $P(x) = mx^3-(2n+4)x^2+4x+m-3=0$ <br\/><b> \u0110\u00e1p s\u1ed1: <\/b> $(m;n)=$(_input_;_input_) ","explain":"<span class='basic_left'> Ta c\u00f3: $P(x)$ chia h\u1ebft cho $x^2+2x = x(x+2)$ n\u00ean $P(x)$ chia h\u1ebft cho $x$ v\u00e0 $x+2$ <br\/> $\\Rightarrow \\left\\{ \\begin{align} & P(0)=0 \\\\ & P(-2)=0 \\\\ \\end{align} \\right.$ $\\Leftrightarrow \\left\\{ \\begin{align} & m-3=0 \\\\ & -8m-4\\left( 2n+4 \\right)-8+m-3=0 \\\\ \\end{align} \\right.$ <br\/> $\\Leftrightarrow \\left\\{ \\begin{align} & m=3 \\\\ & -24-8n-16-8=0 \\\\ \\end{align} \\right.$ $\\Leftrightarrow \\left\\{ \\begin{align} & m=3 \\\\ & 8n=-48 \\\\ \\end{align}\\right.$ $\\Leftrightarrow \\left\\{ \\begin{align} & m=3 \\\\ & n=-6 \\\\ \\end{align} \\right.$ <br\/> <span class='basic_pink'> V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u1ea7n l\u01b0\u1ee3t l\u00e0 $3$ v\u00e0 $-6$<\/span>"}]}],"id_ques":298},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"select":["A. $-\\dfrac{1}{6}$","B. $-\\dfrac{7}{6}$","C. $-\\dfrac{5}{6}$"],"ques":" <span class='basic_left'>X\u00e1c \u0111\u1ecbnh $a$ \u0111\u1ec3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh sau c\u00f3 nghi\u1ec7m: $\\left\\{ \\begin{align} & 3x-y=-16 \\\\ & -2x+ay=a-1 \\\\ & 2x+y=1 \\\\ \\end{align} \\right.$ <br\/>","hint":"X\u00e9t h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh:(I) $\\left\\{ \\begin{align} & 3x-y=-16 \\\\ & 2x+y=1 \\\\ \\end{align} \\right.$<br\/>H\u1ec7 \u0111\u00e3 cho c\u00f3 nghi\u1ec7m n\u1ebfu nghi\u1ec7m c\u1ee7a h\u1ec7 (I) l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $-2x+ay=a-1 $","explain":"<span class='basic_left'> X\u00e9t h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: $\\left\\{ \\begin{align} & 3x-y=-16 \\\\ & 2x+y=1 \\\\ \\end{align} \\right.$ (I)<br\/> \u00c1p d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p c\u1ed9ng \u0111\u1ea1i s\u1ed1, ta \u0111\u01b0\u1ee3c: <br\/> $\\left\\{ \\begin{align} & 3x-y=-16 \\\\ & 2x+y=1 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & 5x=-15 \\\\ & 2x+y=1 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & x=-3 \\\\ & 2.\\left( -3 \\right)+y=1 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & x=-3 \\\\ & y=7 \\\\ \\end{align} \\right.$ <br\/> Suy ra $(-3;7)$ l\u00e0 nghi\u1ec7m c\u1ee7a h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh (I) <br\/> \u0110\u1ec3 h\u1ec7 ba ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m th\u00ec $(-3;7)$ c\u0169ng l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $-2x+ay=a-1$ (*) <br\/> Thay $x= -3; y=7$ v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh (*), ta \u0111\u01b0\u1ee3c: <br\/> $6 + 7a = a -1 \\Leftrightarrow 6a =- 7 \\Leftrightarrow a=\\dfrac{-7}{6}$ <br\/> "}]}],"id_ques":299},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","temp":"fill_the_blank","correct":[[["1"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" T\u00ecm gi\u00e1 tr\u1ecb c\u1ee7a $m$ \u0111\u1ec3 nghi\u1ec7m c\u1ee7a h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh <br\/> $\\left\\{ \\begin{align} & \\dfrac{x+1}{3}-\\dfrac{y+2}{4}=\\dfrac{2\\left( x-y \\right)}{5} \\\\ & \\dfrac{x-3}{4}-\\dfrac{y-3}{3}=2y-x \\\\ \\end{align} \\right.$ <br\/> c\u0169ng l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $3mx - 5y = 2m + 1$<\/b> <br\/><b> \u0110\u00e1p s\u1ed1: <\/b> $m = $ _input_ ","explain":"<span class='basic_left'> X\u00e9t h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: $\\left\\{ \\begin{align} & \\dfrac{x+1}{3}-\\dfrac{y+2}{4}=\\dfrac{2\\left( x-y \\right)}{5} \\\\ & \\dfrac{x-3}{4}-\\dfrac{y-3}{3}=2y-x \\\\ \\end{align} \\right.$ (I) <br\/> R\u00fat g\u1ecdn t\u1eebng ph\u01b0\u01a1ng tr\u00ecnh ta \u0111\u01b0\u1ee3c: <br\/> $ \\left\\{ \\begin{align} & 20\\left( x+1 \\right)-15\\left( y+2 \\right)=24\\left( x-y \\right) \\\\ & 3\\left( x-3 \\right)-4\\left( y-3 \\right)=12\\left( 2y-x \\right) \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & 4x-9y=-10 \\\\ & 15x-28y=-3 \\\\ \\end{align} \\right.$ <br\/> \u00c1p d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p c\u1ed9ng \u0111\u1ea1i s\u1ed1, ta c\u00f3: <br\/> $\\left\\{ \\begin{align} & 4x-9y=-10 \\\\ & 15x-28y=-3 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & 60x-135y=-150 \\\\ & 60x-112y=-12 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & 23y=138 \\\\ & 4x-9y=-10 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & y=6 \\\\ & 4x-9.6=-10 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & 4x=44 \\\\ & y=6 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & x=11 \\\\ & y=6 \\\\ \\end{align} \\right.$ <br\/> Suy ra $(11; 6)$ l\u00e0 nghi\u1ec7m c\u1ee7a h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh (I) <br\/> Do $(11; 6)$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $3mx - 5y = 2m + 1$ (1)<br\/> N\u00ean thay $x= 11; y=6$ v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh (1), ta \u0111\u01b0\u1ee3c: <br\/> $33m -30 = 2m + 1$ $\\Leftrightarrow 31m = 31 \\Leftrightarrow m = 1$<br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $1$ <\/span>"}]}],"id_ques":300}],"lesson":{"save":0,"level":3}}