{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"<span class='basic_left'> Cho $x-y=1$. Gi\u00e1 tr\u1ecb c\u1ee7a \u0111a th\u1ee9c <br\/> $x^3-x^2y-x^2+xy^2-y^3-y^2+5x-5y-5-2012$ l\u00e0: <\/span>","select":["A. $-2011$","B. $-2012$","C. $-2013$ ","D. $-2014$"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Nh\u00f3m $3$ h\u1ea1ng t\u1eed li\u00ean ti\u1ebfp th\u00e0nh m\u1ed9t nh\u00f3m v\u00e0 \u0111\u1eb7t nh\u00e2n t\u1eed chung.<br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3: <br\/> $\\begin{align} & Q={{x}^{3}}-{{x}^{2}}y-{{x}^{2}}+x{{y}^{2}}-{{y}^{3}}-{{y}^{2}}+5x-5y-5-2012 \\\\ & \\,\\,\\,\\,\\,=\\left( {{x}^{3}}-{{x}^{2}}y-{{x}^{2}} \\right)+\\left( x{{y}^{2}}-{{y}^{3}}-{{y}^{2}} \\right)+\\left( 5x-5y-5 \\right)-2012 \\\\ & \\,\\,\\,\\,\\,={{x}^{2}}\\left( x-y-1 \\right)+{{y}^{2}}\\left( x-y-1 \\right)+5\\left( x-y-1 \\right)-2012 \\\\ & \\,\\,\\,\\,\\,={{x}^{2}}\\left( 1-1 \\right)+{{y}^{2}}\\left( 1-1 \\right)+5\\left( 1-1 \\right)-2012 \\\\ & \\,\\,\\,\\,\\,=-2012 \\\\ \\end{align}$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span> <\/span>","column":4}]}],"id_ques":1191},{"time":24,"part":[{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0111\u00e2y \u0110\u00fang hay Sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":" <span class='basic_left'> Cho c\u00e1c \u0111a th\u1ee9c sau: <br\/> $P=5{{x}^{2}}{{y}^{2}}-xy-2{{y}^{3}}-{{y}^{2}}+5{{x}^{4}} $ <br\/> $Q= -2{{x}^{2}}{{y}^{2}}-5xy+{{y}^{3}}-3{{y}^{2}}+2{{x}^{4}} $ <br\/> $N= -{{x}^{2}}{{y}^{2}}+6xy+{{y}^{3}}+6{{y}^{2}}+7$ <br\/> V\u1edbi m\u1ecdi $x;y$ th\u00ec \u00edt nh\u1ea5t m\u1ed9t trong ba \u0111a th\u1ee9c tr\u00ean c\u00f3 gi\u00e1 tr\u1ecb d\u01b0\u01a1ng. <\/span>","select":["\u0110\u00fang","Sai "],"hint":"So s\u00e1nh t\u1ed5ng $P+Q+N$ v\u1edbi $0$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>X\u00e9t t\u1ed5ng ba \u0111a th\u1ee9c \u0111\u00e3 cho v\u00e0 so s\u00e1nh t\u1ed5ng \u0111\u00f3 v\u1edbi $0$ <br\/> N\u1ebfu $P+Q+N > 0$ th\u00ec kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 \u0111\u00fang. <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> X\u00e9t t\u1ed5ng ba \u0111a th\u1ee9c \u0111\u00e3 cho: <br\/> $\\begin{aligned} & P+Q+N \\\\ & =\\left( 5{{x}^{2}}{{y}^{2}}-xy-2{{y}^{3}}-{{y}^{2}}+5{{x}^{4}} \\right)+\\left( -2{{x}^{2}}{{y}^{2}}-5xy+{{y}^{3}}-3{{y}^{2}}+2{{x}^{4}} \\right)+\\left( -{{x}^{2}}{{y}^{2}}+6xy+{{y}^{3}}+6{{y}^{2}}+7 \\right) \\\\ & =5{{x}^{2}}{{y}^{2}}-xy-2{{y}^{3}}-{{y}^{2}}+5{{x}^{4}}-2{{x}^{2}}{{y}^{2}}-5xy+{{y}^{3}}-3{{y}^{2}}+2{{x}^{4}}-{{x}^{2}}{{y}^{2}}+6xy+{{y}^{3}}+6{{y}^{2}}+7 \\\\ & =\\left( 5{{x}^{2}}{{y}^{2}}-2{{x}^{2}}{{y}^{2}}-{{x}^{2}}{{y}^{2}} \\right)+\\left( -xy-5xy+6xy \\right)+\\left( -2{{y}^{3}}+{{y}^{3}}+{{y}^{3}} \\right)+\\left( -{{y}^{2}}-3{{y}^{2}}+6{{y}^{2}} \\right)+\\left( 5{{x}^{4}}+2{{x}^{4}} \\right)+7 \\\\ & =2{{x}^{2}}{{y}^{2}}+2{{y}^{2}}+7{{x}^{4}}+7 \\\\ & Do\\,\\,\\left\\{ \\begin{aligned} & 2{{x}^{2}}{{y}^{2}}\\ge 0 \\\\ & 2{{y}^{2}}\\ge 0 \\\\ & 7{{x}^{4}}\\ge 0 \\\\ \\end{aligned} \\right.\\Rightarrow 2{{x}^{2}}{{y}^{2}}+2{{y}^{2}}+7{{x}^{4}}+7>0\\,\\,\\,\\,\\forall x;y \\\\ & \\, \\\\ \\end{aligned}$ <br\/> $\\Rightarrow$ V\u1edbi m\u1ecdi $x;y$ th\u00ec \u00edt nh\u1ea5t m\u1ed9t trong ba \u0111a th\u1ee9c tr\u00ean c\u00f3 gi\u00e1 tr\u1ecb d\u01b0\u01a1ng. <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang.<\/span><\/span>","column":2}]}],"id_ques":1192},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"<span class='basic_left'> Cho hai \u0111a th\u1ee9c sau: <br\/> $P=\\left[ \\dfrac{1}{2}a\\,x-2\\left( a\\,x+3 \\right) \\right]-\\left( a\\,x+1 \\right)$ <br\/> $Q=a\\,x-2-\\left[ 3-\\left( a\\,x-1 \\right) \\right]-4$ <br\/> T\u1ed5ng $P+Q$ l\u00e0: <\/span>","select":["A. $\\dfrac{3}{2}a\\,x+16$","B. $-\\dfrac{3}{2}a\\,x-16$","C. $\\dfrac{1}{2}a\\,x+17$ ","D. $-\\dfrac{1}{2}a\\,x-17$"],"explain":"<span class='basic_left'>Ta c\u00f3: <br\/> $\\begin{align} & P+Q \\\\ & =\\left[ \\dfrac{1}{2}a\\,x-2\\left( a\\,x+3 \\right) \\right]-\\left( a\\,x+1 \\right)+a\\,x-2-\\left[ 3-\\left( a\\,x-1 \\right) \\right]-4 \\\\ & =\\dfrac{1}{2}a\\,x-2a\\,x-6-a\\,x-1+a\\,x-2-3+a\\,x-1-4 \\\\ & =\\left( \\dfrac{1}{2}a\\,x-2a\\,x-a\\,x+a\\,x+a\\,x \\right)+\\left( -6-1-2-3-1-4 \\right) \\\\ & =-\\dfrac{1}{2}a\\,x-17 \\\\ \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span> <\/span>","column":2}]}],"id_ques":1193},{"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":"<span class='basic_left'> T\u1ed5ng c\u1ee7a \u0111a th\u1ee9c $A$ v\u1edbi \u0111a th\u1ee9c $2x^4-3x^2y+y^4+3xz+z^2$ l\u00e0 m\u1ed9t \u0111a th\u1ee9c kh\u00f4ng ch\u1ee9a bi\u1ebfn $x.$ <br\/> \u0110a th\u1ee9c $A$ th\u1ecfa m\u00e3n \u0111\u1ec1 b\u00e0i l\u00e0: <\/span>","select":["A. $-2x^4+3x^2y-3xz$","B. $2x^4+3x^2y-3xz+1$","C. $x^4+x^2y-xz$ ","D. $-x^4-x^2y+xz+1$"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> \u0110\u1ec3 t\u1ed5ng c\u1ee7a $A$ v\u00e0 \u0111a th\u1ee9c \u0111\u00e3 cho l\u00e0 m\u1ed9t \u0111a th\u1ee9c kh\u00f4ng ch\u1ee9a bi\u1ebfn $x$ th\u00ec \u0111a th\u1ee9c $A$ c\u00f3 nhi\u1ec1u \u0111\u00e1p \u00e1n th\u1ecfa m\u00e3n. <br\/> Do \u0111\u00f3 ta th\u1eed \u0111a th\u1ee9c $A$ \u1edf c\u00e1c \u0111\u00e1p \u00e1n xem \u0111\u00e1p \u00e1n n\u00e0o th\u1ecfa m\u00e3n.<br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3: <br\/> $\\begin{align} & \\left( -2{{x}^{4}}+3{{x}^{2}}y-3xz \\right)+\\left( 2{{x}^{4}}-3{{x}^{2}}y+{{y}^{4}}+3xz+{{z}^{2}} \\right) \\\\ & =\\left( -2{{x}^{4}}+2{{x}^{4}} \\right)+\\left( 3{{x}^{2}}y-3{{x}^{2}}y \\right)+\\left( -3xz+3xz \\right)+{{y}^{4}}+{{z}^{2}} \\\\ & ={{y}^{4}}+{{z}^{2}}\\,\\, \\\\ \\end{align}$ <br\/> $\\Rightarrow$ \u0110\u00e1p \u00e1n A th\u1ecfa m\u00e3n. <br\/> $\\begin{align} & \\left( 2{{x}^{4}}+3{{x}^{2}}y-3xz+1 \\right)+\\left( 2{{x}^{4}}-3{{x}^{2}}y+{{y}^{4}}+3xz+{{z}^{2}} \\right) \\\\ & =\\left( 2{{x}^{4}}+2{{x}^{4}} \\right)+\\left( 3{{x}^{2}}y-3{{x}^{2}}y \\right)+\\left( -3xz+3xz \\right)+1+{{y}^{4}}+{{z}^{2}} \\\\ & =4{{x}^{4}}+1+{{y}^{4}}+{{z}^{2}}\\,\\, \\\\ \\end{align}$ <br\/> $\\Rightarrow$ \u0110\u00e1p \u00e1n B kh\u00f4ng th\u1ecfa m\u00e3n. <br\/> $\\begin{align} & \\left( {{x}^{4}}+{{x}^{2}}y-xz \\right)+\\left( 2{{x}^{4}}-3{{x}^{2}}y+{{y}^{4}}+3xz+{{z}^{2}} \\right) \\\\ & =\\left( {{x}^{4}}+2{{x}^{4}} \\right)+\\left( {{x}^{2}}y-3{{x}^{2}}y \\right)+\\left( -xz+3xz \\right)+{{y}^{4}}+{{z}^{2}} \\\\ & =3{{x}^{4}}-2{{x}^{2}}y+2xz+{{y}^{4}}+{{z}^{2}}\\,\\, \\\\ \\end{align}$ <br\/>$\\Rightarrow$ \u0110\u00e1p \u00e1n C kh\u00f4ng th\u1ecfa m\u00e3n. <br\/> $\\begin{align}& \\left( -{{x}^{4}}-{{x}^{2}}y+xz+1 \\right)+\\left( 2{{x}^{4}}-3{{x}^{2}}y+{{y}^{4}}+3xz+{{z}^{2}} \\right) \\\\ & =\\left( -{{x}^{4}}+2{{x}^{4}} \\right)+\\left( -{{x}^{2}}y-3{{x}^{2}}y \\right)+\\left( xz+3xz \\right)+1+{{y}^{4}}+{{z}^{2}} \\\\ & ={{x}^{4}}-4{{x}^{2}}y+4xz+1+{{y}^{4}}+{{z}^{2}}\\,\\, \\\\ \\end{align}$ <br\/> $\\Rightarrow $ \u0110\u00e1p \u00e1n D kh\u00f4ng th\u1ecfa m\u00e3n. <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span> <\/span>","column":2}]}],"id_ques":1194},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 \u0111\u00fang nh\u1ea5t v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["1"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":" <span class='basic_left'> Bi\u1ebft $x+y-2=0.$ <br\/> Gi\u00e1 tr\u1ecb c\u1ee7a \u0111a th\u1ee9c $M=x^3+x^2y-2x^2-xy-y^2+3y+x-1$ l\u00e0 _input_ <\/span>","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Bi\u1ebfn \u0111\u1ed5i $M$ l\u00e0m xu\u1ea5t hi\u1ec7n th\u1eeba s\u1ed1 $x+y-2$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3: <br\/> $\\begin{align} & M={{x}^{3}}+{{x}^{2}}y-2{{x}^{2}}-xy-{{y}^{2}}+3y+x-1 \\\\ & \\,\\,\\,\\,\\,\\,=\\left( {{x}^{3}}+{{x}^{2}}y-2{{x}^{2}} \\right)+\\left( -xy-{{y}^{2}}+2y \\right)+\\left( x+y-2 \\right)+1 \\\\ & \\,\\,\\,\\,\\,\\,={{x}^{2}}\\left( x+y-2 \\right)-y\\left( x+y-2 \\right)+\\left( x+y-2 \\right)+1 \\\\ & \\,\\,\\,\\,\\,={{x}^{2}}.0-y.0+0+1 \\\\ & \\,\\,\\,\\,\\,=1 \\\\ \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $1$ <\/span><\/span>"}]}],"id_ques":1195},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 \u0111\u00fang nh\u1ea5t v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["3"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":" <span class='basic_left'> T\u00ecm $x,$ bi\u1ebft $|x-2|=|4-x|$ <br\/> <b> \u0110\u00e1p \u00e1n:<\/b> $x=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$ <\/span>","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> $|A|=|B| \\Rightarrow \\left[ \\begin{align} & A=B \\\\ & A=-B \\\\ \\end{align} \\right.$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3: <br\/> $\\begin{aligned} & |x-2|=|4-x| \\\\ & \\Rightarrow \\left[ \\begin{aligned} & x-2=4-x \\\\ & x-2=x-4 \\\\ \\end{aligned} \\right. \\\\ & \\Rightarrow \\left[ \\begin{aligned} & 2x=6 \\\\ & 0=-2\\,\\text{(v\u00f4 l\u00ed)} \\\\ \\end{aligned} \\right. \\\\ & \\Rightarrow x=3 \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $3$ <\/span><\/span>"}]}],"id_ques":1196},{"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":[[3]],"list":[{"point":10,"ques":"<span class='basic_left'> Cho bi\u1ebft $M+\\left( 3{{x}^{3}}+3{{x}^{2}}y-3x{{y}^{2}}+xy-{{x}^{2}}-1 \\right)=\\left( 3{{x}^{3}}+3{{x}^{2}}y-3x{{y}^{2}}+xy \\right)$ <br\/> <b> C\u00e2u 1:<\/b> \u0110a th\u1ee9c $M$ l\u00e0: <\/span>","select":["A. $2xy+x+1$","B. $6x^2y+6xy^2+x^2+1$","C. $x^2+1$ ","D. $1$"],"explain":"<span class='basic_left'> Ta c\u00f3: <br\/> $\\begin{align} & M+\\left( 3{{x}^{3}}+3{{x}^{2}}y-3x{{y}^{2}}+xy-{{x}^{2}}-1 \\right)=\\left( 3{{x}^{3}}+3{{x}^{2}}y-3x{{y}^{2}}+xy \\right) \\\\ & M=\\left( 3{{x}^{3}}+3{{x}^{2}}y-3x{{y}^{2}}+xy \\right)-\\left( 3{{x}^{3}}+3{{x}^{2}}y-3x{{y}^{2}}+xy-{{x}^{2}}-1 \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,=3{{x}^{3}}+3{{x}^{2}}y-3x{{y}^{2}}+xy-3{{x}^{3}}-3{{x}^{2}}y+3x{{y}^{2}}-xy+{{x}^{2}}+1 \\\\ & \\,\\,\\,\\,\\,\\,\\,=\\left( 3{{x}^{3}}-3{{x}^{3}} \\right)+\\left( 3{{x}^{2}}y-3{{x}^{2}}y \\right)+\\left( -3x{{y}^{2}}+3x{{y}^{2}} \\right)+\\left( xy-xy \\right)+{{x}^{2}}+1 \\\\ & \\,\\,\\,\\,\\,\\,\\,=0+0+0+0+{{x}^{2}}+1 \\\\ & \\,\\,\\,\\,\\,\\,\\,={{x}^{2}}+1 \\\\ \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span> <\/span>","column":2}]}],"id_ques":1197},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 \u0111\u00fang nh\u1ea5t v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"fill_the_blank_random","correct":[[["4"],["-4"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":" <span class='basic_left'> Cho bi\u1ebft $M+\\left( 3{{x}^{3}}+3{{x}^{2}}y-3x{{y}^{2}}+xy-{{x}^{2}}-1 \\right)=\\left( 3{{x}^{3}}+3{{x}^{2}}y-3x{{y}^{2}}+xy \\right)$ <br\/> <b> C\u00e2u 2:<\/b> T\u00ecm gi\u00e1 tr\u1ecb c\u1ee7a $x$ \u0111\u1ec3 $M=17$ <br\/> <b> \u0110\u00e1p \u00e1n:<\/b> $\\left[ \\begin{align} & x=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & x=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ \\end{align} \\right.$ <\/span>","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Cho \u0111a th\u1ee9c $M$ b\u1eb1ng $17$ v\u00e0 t\u00ecm $x$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Theo c\u00e2u 1, ta t\u00ecm \u0111\u01b0\u1ee3c $M=x^2+1$ <br\/> Ta c\u00f3: <br\/> $x^2+1=17$ <br\/> $x^2=17-1$ <br\/> $x^2=16$ <br\/> $\\Rightarrow \\left[ \\begin{align} & x=4 \\\\ & x=-4 \\\\ \\end{align} \\right.$ <br\/> <span class='basic_pink'> V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o c\u00e1c \u00f4 tr\u1ed1ng l\u00e0 $4$ v\u00e0 $-4$ <\/span><\/span>"}]}],"id_ques":1198},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 \u0111\u00fang nh\u1ea5t v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["1"],["1"],["1"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":" <span class='basic_left'> T\u00ecm gi\u00e1 tr\u1ecb c\u1ee7a $x;y;z$ th\u1ecfa m\u00e3n $(1-x)^2+(x-y)^2+(y-z)^2=0$ <br\/> <b> \u0110\u00e1p \u00e1n:<\/b> $\\left\\{ \\begin{align} & x=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & y=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & z=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ \\end{align} \\right. $ <\/span>","explain":"<span class='basic_left'> $\\begin{aligned} & Do\\,\\,{{\\left( 1-x \\right)}^{2}}\\ge 0;\\,\\,{{\\left( x-y \\right)}^{2}}\\,\\,\\ge 0;\\,\\,{{\\left( y-z \\right)}^{2}}\\,\\ge 0 \\,\\forall\\,x,y,z \\\\ & \u0110\u1ec3\\,\\,{{(1-x)}^{2}}+{{(x-y)}^{2}}+{{(y-z)}^{2}}=0 \\\\ & \\Rightarrow \\left\\{ \\begin{aligned} & 1-x=0 \\\\ & x-y=0 \\\\ & y-z=0 \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & x=1 \\\\ & y=x=1 \\\\ & z=y=1 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'> V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o c\u00e1c \u00f4 tr\u1ed1ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 $1;1$ v\u00e0 $1$ <\/span><\/span>"}]}],"id_ques":1199},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 \u0111\u00fang nh\u1ea5t v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["6"],["0"],["0"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":" <span class='basic_left'> T\u00ecm gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a bi\u1ec3u th\u1ee9c: <br\/> $M=\\left( {{x}^{2}}{{y}^{3}}+{{x}^{3}}{{y}^{2}}-{{x}^{2}}+{{y}^{2}}+5 \\right)-\\left( {{x}^{2}}{{y}^{3}}+{{x}^{3}}{{y}^{2}}+2{{y}^{2}}-1 \\right)$ <br\/> <b> \u0110\u00e1p \u00e1n:<\/b> Gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a $M$ l\u00e0 _input_ khi $\\left\\{ \\begin{align} & x=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & y=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ \\end{align} \\right. $ <\/span>","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Thu g\u1ecdn bi\u1ec3u th\u1ee9c $M$ <br\/> Nh\u1eadn \u0111\u1ecbnh, \u0111\u00e1nh gi\u00e1 \u0111\u1ec3 t\u00ecm $Max\\,M$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> $\\begin{aligned} & M=\\left( {{x}^{2}}{{y}^{3}}+{{x}^{3}}{{y}^{2}}-{{x}^{2}}+{{y}^{2}}+5 \\right)-\\left( {{x}^{2}}{{y}^{3}}+{{x}^{3}}{{y}^{2}}+2{{y}^{2}}-1 \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,={{x}^{2}}{{y}^{3}}+{{x}^{3}}{{y}^{2}}-{{x}^{2}}+{{y}^{2}}+5-{{x}^{2}}{{y}^{3}}-{{x}^{3}}{{y}^{2}}-2{{y}^{2}}+1 \\\\ & \\,\\,\\,\\,\\,\\,\\,=\\left( {{x}^{2}}{{y}^{3}}-{{x}^{2}}{{y}^{3}} \\right)+\\left( {{x}^{3}}{{y}^{2}}-{{x}^{3}}{{y}^{2}} \\right)-{{x}^{2}}+\\left( {{y}^{2}}-2{{y}^{2}} \\right)+5+1 \\\\ & \\,\\,\\,\\,\\,\\,\\,=0+0-{{x}^{2}}-{{y}^{2}}+6 \\\\ & \\,\\,\\,\\,\\,\\,\\,=-{{x}^{2}}-{{y}^{2}}+6 \\\\ & Do\\,\\,\\left\\{ \\begin{aligned} & {{x}^{2}}\\,\\,\\ge 0\\,\\forall\\,x \\\\ & {{y}^{2}}\\ge 0\\,\\forall\\,y \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & -{{x}^{2}}\\le 0 \\,\\forall\\,x \\\\ & -{{y}^{2}}\\le 0 \\,\\forall \\,y \\\\ \\end{aligned} \\right. \\\\ & \\Rightarrow M=-{{x}^{2}}-{{y}^{2}}+6\\le 6 \\,\\forall\\,x,y \\\\ & \\Rightarrow Ma{{x}{M}}=6\\,\\,khi\\,\\,x=y=0 \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'> V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o c\u00e1c \u00f4 tr\u1ed1ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 $6;0$ v\u00e0 $0$ <\/span><\/span>"}]}],"id_ques":1200}],"lesson":{"save":0,"level":3}}