{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng trong ph\u00e9p chia sau ","title_trans":"","temp":"fill_the_blank","correct":[[["-100"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai10/lv3/img\/16.jpg' \/><\/center> Gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a bi\u1ec3u th\u1ee9c $C=\\left( x-2 \\right)\\left( x-5 \\right)\\left( {{x}^{2}}-7x-10 \\right)$ l\u00e0 _input_ ","hint":"Ph\u00e2n t\u00edch $C$, \u0111\u01b0a bi\u1ec3u th\u1ee9c v\u1ec1 d\u1ea1ng $[f(x)]^2+a$ v\u1edbi $a$ l\u00e0 h\u1eb1ng s\u1ed1. <br\/> GTNN c\u1ee7a $C$ l\u00e0 $a$, \u0111\u1ea1t \u0111\u01b0\u1ee3c khi $[f(x)]^2 = 0$ (t\u00ecm $x$ \u0111\u1ec3 $[f(x)]^2 = 0$).","explain":" <span class='basic_left'>Ta c\u00f3: <br\/> $C=\\left( x-2 \\right)\\left( x-5 \\right)\\left( {{x}^{2}}-7x-10 \\right) \\\\ =\\left( {{x}^{2}}-7x+10 \\right)\\left( {{x}^{2}}-7x-10 \\right) \\\\ ={{\\left( {{x}^{2}}-7x \\right)}^{2}}-100$<br\/> Do ${{\\left( {{x}^{2}}-7x \\right)}^{2}}\\ge 0$$\\Rightarrow {{\\left( {{x}^{2}}-7x \\right)}^{2}}-100\\ge -100$ v\u1edbi $\\forall x$ <br\/> V\u1eady GTNN c\u1ee7a $C$ l\u00e0 $\u2013 100$ . <br\/> D\u1ea5u \u201c$=$\u201d x\u1ea3y ra khi ${{\\left( {{x}^{2}}-7x \\right)}^{2}}=0 \\Leftrightarrow {{x}^{2}}-7x=0 \\Leftrightarrow \\left[ \\begin{align} & x=0 \\\\ & x=7 \\\\ \\end{align} \\right.$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-100$. <\/span><\/span><\/span> "}]}],"id_ques":771},{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai10/lv3/img\/8.jpg' \/><\/center> Ph\u00e9p chia $({{x}^{27}}+{{x}^{9}}+{{x}^{3}}+x):(x^2-1)$ c\u00f3 \u0111a th\u1ee9c d\u01b0 l\u00e0: ","select":["A. $4x$ ","B. $3x$","C. $x+2$","D. $x-2$"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1:<\/b> T\u00e1ch v\u00e0 nh\u00f3m $({{x}^{27}}+{{x}^{9}}+{{x}^{3}}+x)=\\left( {{x}^{27}}-x \\right)+\\left( {{x}^{9}}-x \\right)+\\left( {{x}^{3}}-x \\right)+4x$<br\/> <b> B\u01b0\u1edbc 2:<\/b> Ph\u00e2n t\u00edch \u0111a th\u1ee9c b\u1eb1ng c\u00e1ch \u0111\u1eb7t $x$ l\u00e0m nh\u00e2n t\u1eed chung. <br\/> <b> B\u01b0\u1edbc 3:<\/b> \u0110\u00e1nh gi\u00e1 \u0111a th\u1ee9c chia cho $x^2-1$ c\u00f3 d\u01b0 l\u00e0 \u0111a th\u1ee9c n\u00e0o.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> $\\begin{align} & {{x}^{27}}+{{x}^{9}}+{{x}^{3}}+x \\\\ & =\\left( {{x}^{27}}-x \\right)+\\left( {{x}^{9}}-x \\right)+\\left( {{x}^{3}}-x \\right)+4x \\\\ & = x \\left[ \\left( {{x}^{26}}-1 \\right)+\\left( {{x}^{8}}-1 \\right)+\\left( {{x}^{2}}-1 \\right) \\right]+4x \\\\ & \\text{Do} \\,\\,\\,\\left. \\begin{aligned} & \\left( {{x}^{26}}-1 \\right)\\,\\,\\vdots \\,\\,\\,\\left( {{x}^{2}}-1 \\right) \\\\ & \\left( {{x}^{8}}-1 \\right)\\,\\vdots \\,\\,\\,\\left( {{x}^{2}}-1 \\right) \\\\ & \\left( {{x}^{2}}-1 \\right)\\,\\vdots \\,\\,\\,\\left( {{x}^{2}}-1 \\right) \\\\ \\end{aligned} \\right\\}\\\\ & \\Rightarrow \\left[ \\left( {{x}^{26}}-1 \\right)+\\left( {{x}^{8}}-1 \\right)+\\left( {{x}^{2}}-1 \\right) \\right]\\,\\,\\vdots \\,\\,\\,\\left( {{x}^{2}}-1 \\right) \\\\ & \\Rightarrow x \\left[ \\left( {{x}^{26}}-1 \\right)+\\left( {{x}^{8}}-1 \\right)+\\left( {{x}^{2}}-1 \\right) \\right]\\,\\vdots \\,\\,\\,\\left( {{x}^{2}}-1 \\right) \\\\ \\end{align}$ <br\/> Nh\u01b0ng $4x\\,\\not\\vdots\\, (x^2-1)$ <br\/>V\u1eady $\\left( {{x}^{27}}+{{x}^{9}}+{{x}^{3}}+x \\right)\\,\\,:\\,\\,\\left( {{x}^{2}}-1 \\right)\\,\\,$ c\u00f3 \u0111a th\u1ee9c d\u01b0 l\u00e0 $4x$. <br\/> <span class='basic_pink'> V\u1eady ch\u1ecdn \u0111\u00e1p \u00e1n A. <\/span><\/span> ","column":2}]}],"id_ques":772},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank","correct":[[["0"],["-1"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai10/lv3/img\/5.jpg' \/><\/center>T\u00ecm $m, n$ $( m, n\\,\\in\\,\\mathbb{Z})$ \u0111\u1ec3 ph\u00e9p chia sau l\u00e0 ph\u00e9p chia h\u1ebft:<br\/> $(x^4+mx+n):(x^2-1)$<br\/> \u0110\u00e1p \u00e1n: $\\begin{cases} m &= \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\\\ n& =\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\end{cases} $ ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> \u0110\u1eb7t ph\u00e9p chia $(x^4+mx+n):(x^2-1)$ t\u00ecm s\u1ed1 d\u01b0, cho s\u1ed1 d\u01b0 b\u1eb1ng 0 \u0111\u1ec3 t\u00ecm m, n.<\/b> <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> $\\left. \\begin{align} & \\begin{matrix} {{x}^{4}}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,+mx+n\\,\\, \\\\ {{x}^{4}}-{{x}^{2}}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\\\\\end{matrix} \\\\ & \\overline{\\begin{align} & \\,\\,\\,\\,\\,\\,\\,\\,{{x}^{2}}+mx+n\\,\\,\\,\\,\\, \\\\ & \\,\\,\\,\\,\\,\\,\\,{{x}^{2}}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-1\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\overline{\\,\\,\\,\\,\\,\\,\\,\\,\\,mx+n+1\\,\\,\\,\\,} \\\\ \\end{align}} \\\\ \\end{align} \\right|\\begin{matrix} \\dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} \\\\ \\begin{matrix} {} \\\\ {} \\\\\\end{matrix} \\\\ {} \\\\\\end{matrix}$ <br\/> V\u1eady $(x^4+mx+n):(x^2-1)=x^2+1$ d\u01b0 $mx+n+1$<br\/> \u0110\u1ec3 $(x^4+mx+n):(x^2-1)$ l\u00e0 ph\u00e9p chia h\u1ebft th\u00ec $ mx+n+1=0 \\Leftrightarrow \\begin{cases} m&=0 \\\\ n&=-1 \\end{cases} $. <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $0$ v\u00e0 $-1$. <\/span><\/span> "}]}],"id_ques":773},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng trong ph\u00e9p chia sau ","title_trans":"","temp":"fill_the_blank","correct":[[["3x-2"],["3x+3","3+3x","3(x+1)","3(1+x)"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai10/lv3/img\/3.jpg' \/><\/center> $(3x^3-2x^2+5):(x^2-1)=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$ d\u01b0 $\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> \u0110\u1eb7t ph\u00e9p chia $(3x^3-2x^2+5):(x^2-1)$ t\u00ecm th\u01b0\u01a1ng v\u00e0 s\u1ed1 d\u01b0. <\/b> <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> $\\left. \\begin{align} & \\begin{matrix} 3{{x}^{3}}-2{{x}^{2}}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,+5 \\\\ 3{{x}^{3}}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-3x\\,\\,\\,\\,\\,\\,\\,\\, \\\\\\end{matrix} \\\\ & \\overline{\\begin{align} & \\,\\,\\,\\,\\,\\,\\,-2{{x}^{2}}+3x+5\\,\\,\\,\\,\\, \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,-2{{x}^{2}}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,+2\\,\\, \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\overline{\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,3x+3\\,\\,\\,\\,\\,\\,\\,\\,} \\\\ \\end{align}} \\\\ \\end{align} \\right|\\begin{matrix} \\dfrac{{{x}^{2}}-1}{3x-2} \\\\ \\begin{matrix} {} \\\\{} \\\\\\end{matrix} \\\\ {} \\\\\\end{matrix}$ <br\/> V\u1eady $(3x^3-2x^2+5):(x^2-1)=3x-2$ d\u01b0 $3x+3$. <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $3x-2$ v\u00e0 $3x+3$. <\/span><\/span> "}]}],"id_ques":774},{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai10/lv3/img\/13.jpg' \/><\/center> Hi\u1ec7u c\u00e1c b\u00ecnh ph\u01b0\u01a1ng c\u1ee7a hai s\u1ed1 l\u1ebb li\u00ean ti\u1ebfp lu\u00f4n chia h\u1ebft cho ","select":["A. $11$ ","B. $5$","C. $6$","D. $8$"],"hint":"G\u1ecdi hai s\u1ed1 l\u1ebb li\u00ean ti\u1ebfp l\u00e0 $2n+1$ v\u00e0 $2n+3$ v\u1edbi m\u1ecdi $n\\in\\mathbb {N}$.<br\/> Ph\u00e2n t\u00edch $(2n+3)^2-(2n+1)^2$ th\u00e0nh nh\u00e2n t\u1eed theo h\u1eb1ng \u0111\u1eb3ng th\u1ee9c hi\u1ec7u hai b\u00ecnh ph\u01b0\u01a1ng. ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> G\u1ecdi hai s\u1ed1 l\u1ebb li\u00ean ti\u1ebfp l\u00e0 $2n+1$ v\u00e0 $2n+3$ v\u1edbi m\u1ecdi $n\\in\\mathbb {N}$<br\/> <b> B\u01b0\u1edbc 2:<\/b> Ph\u00e2n t\u00edch $(2n+3)^2-(2n+1)^2$ th\u00e0nh nh\u00e2n t\u1eed theo h\u1eb1ng \u0111\u1eb3ng th\u1ee9c hi\u1ec7u hai b\u00ecnh ph\u01b0\u01a1ng.<br\/> <b> B\u01b0\u1edbc 3:<\/b> Nh\u1eadn \u0111\u1ecbnh bi\u1ec3u th\u1ee9c \u0111\u00e3 cho chia h\u1ebft cho s\u1ed1 n\u00e0o trong c\u00e1c \u0111\u00e1p \u00e1n \u0111\u00e3 cho. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>G\u1ecdi hai s\u1ed1 l\u1ebb li\u00ean ti\u1ebfp l\u00e0 $2n+1$ v\u00e0 $2n+3$ v\u1edbi m\u1ecdi $n\\in\\mathbb {N}$<br\/> Ta c\u00f3:<br\/> $\\begin{align} & (2n+3)^2-(2n+1)^2 \\\\ & =(2n+3+2n+1)(2n+3-2n-1) \\\\ & =(4n+4).2 \\\\ & =8(n+1) \\\\ \\end{align}$ <br\/> Do $8\\,\\vdots\\,8 \\Rightarrow 8(n+1) \\,\\vdots \\,8$. <br\/> <span class='basic_pink'> V\u1eady ch\u1ecdn \u0111\u00e1p \u00e1n D. <\/span><\/span> ","column":2}]}],"id_ques":775},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank","correct":[[["-2"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai10/lv3/img\/11.jpg' \/><\/center> T\u00ecm $x$, bi\u1ebft $(x^4+8x^2+16):(x^2+4)+4x=0$ <br\/> \u0110\u00e1p \u00e1n: $x$ = _input_ ","hint":" Th\u1ef1c hi\u1ec7n ph\u00e9p chia $(x^4+8x^2+16):(x^2+4)$ sau \u0111\u00f3 r\u00fat g\u1ecdn \u0111\u01b0a v\u1ec1 d\u1ea1ng $ax=b$ ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span> <br\/> <b> B\u01b0\u1edbc 1:<\/b> Th\u1ef1c hi\u1ec7n ph\u00e9p chia $(x^4+8x^2+16):(x^2+4)$<br\/> <b> B\u01b0\u1edbc 2:<\/b> R\u00fat g\u1ecdn \u0111\u01b0a v\u1ec1 d\u1ea1ng $ax=b$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> Ta c\u00f3: <br\/> $\\begin{align} & ({{x}^{4}}+8{{x}^{2}}+16):({{x}^{2}}+4)+4x =0 \\\\ &\\Leftrightarrow {{\\left( {{x}^{2}}+4 \\right)}^{2}}:\\left( {{x}^{2}}+4 \\right)+4x =0 \\\\ &\\Leftrightarrow {{x}^{2}}+4+4x =0 \\\\ &\\Leftrightarrow {{x}^{2}}+4x+4 =0 \\\\ &\\Leftrightarrow {{\\left( x+2 \\right)}^{2}} =0 \\\\ &\\Leftrightarrow x+2 =0 \\\\ &\\Leftrightarrow x =-2 \\\\ \\end{align}$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-2$. <\/span><\/span> "}]}],"id_ques":776},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank_random","correct":[[["2"],["3"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai10/lv3/img\/10.jpg' \/><\/center> T\u00ecm $x$, bi\u1ebft $3(2x-3)(3x+2)-2(x+4)(4x-3)+9x(4-x)=0$ <br\/> \u0110\u00e1p \u00e1n: $\\left[ \\begin{align} & x=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & x=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ \\end{align} \\right.$ ","hint":"Bi\u1ebfn \u0111\u1ed5i v\u1ebf tr\u00e1i v\u1ec1 d\u1ea1ng $a.b=0$<br\/> N\u1ebfu $a.b=0$ th\u00ec $\\left[ \\begin{align} & a=0 \\\\ & b=0 \\\\ \\end{align} \\right.$ ","explain":"<span class='basic_left'> Ta c\u00f3: <br\/> $\\begin{aligned} & 3(2x-3)(3x+2)-2(x+4)(4x-3)+9x(4-x)=0 \\\\ &\\Leftrightarrow \\left( 6x-9 \\right)\\left( 3x+2 \\right)-\\left( 2x+8 \\right)\\left( 4x-3 \\right)+36x-9{{x}^{2}}=0 \\\\ &\\Leftrightarrow 18{{x}^{2}}+12x-27x-18-8{{x}^{2}}+6x-32x+24+36x-9{{x}^{2}}=0 \\\\ &\\Leftrightarrow {{x}^{2}}-5x+6=0 \\\\ &\\Leftrightarrow {{x}^{2}}-2x-3x+6=0 \\\\ &\\Leftrightarrow x\\left( x-2 \\right)-3\\left( x-2 \\right)=0 \\\\ &\\Leftrightarrow \\left( x-2 \\right)\\left( x-3 \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x-2=0 \\\\ & x-3=0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left[ \\begin{aligned} & x=2 \\\\ & x=3 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $2$ v\u00e0 $3$. <\/span><\/span> "}]}],"id_ques":777},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank","correct":[[["0"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai10/lv3/img\/8.jpg' \/><\/center> Ph\u00e2n t\u00edch \u0111a th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed v\u00e0 t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $x^3+3x^2+3x+1-27z^3$ t\u1ea1i $x=2;z=1$ l\u00e0 _input_","hint":"\u00c1p d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c l\u1eadp ph\u01b0\u01a1ng c\u1ee7a m\u1ed9t t\u1ed5ng v\u00e0 hi\u1ec7u hai l\u1eadp ph\u01b0\u01a1ng \u0111\u1ec3 ph\u00e2n t\u00edch \u0111a th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span> <br\/> <b> B\u01b0\u1edbc 1:<\/b> Ph\u00e2n t\u00edch t\u00edch \u0111a th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed<br\/> <b> B\u01b0\u1edbc 2:<\/b> Thay $x=2;z=1$ v\u00e0o \u0111\u1ec3 t\u00ednh. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> Ta c\u00f3: <br\/> $\\begin{align} & x^3+3x^2+3x+1-27z^3 \\\\ &=(x+1)^3-(3z)^3 \\\\ & =(x+1-3z)[(x+1)^2+3z(x+1)+9z^2] \\\\ \\end{align}$ <br\/> Thay $x=2;z=1$ v\u00e0o bi\u1ec3u th\u1ee9c , ta \u0111\u01b0\u1ee3c: <br\/> $(2+1-3.1)[(2+1)^2+3.1(2+1)+9.1^2] =0 $ . <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $0$. <\/span><\/span> "}]}],"id_ques":778},{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai10/lv3/img\/4.jpg' \/><\/center> K\u1ebft qu\u1ea3 ph\u00e2n t\u00edch \u0111a th\u1ee9c $p^6q-p^5q^3-p^2q^4+pq^6$ th\u00e0nh nh\u00e2n t\u1eed l\u00e0: ","select":["A. $(p-q)(p-q^2)$ ","B. $pq(p-q^2)(p^4-q^3)$","C. $pq(p-q)$","D. $pq(p-q^2)(p-q^3)$"],"hint":"S\u1eed d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p nh\u00f3m h\u1ea1ng t\u1eed v\u00e0 \u0111\u1eb7t nh\u00e2n t\u1eed chung \u0111\u1ec3 ph\u00e2n t\u00edch.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1:<\/b> Nh\u00f3m $p^6q-p^5q^3-p^2q^4+pq^6=(p^6q-p^5q^3)-(p^2q^4-pq^6)$.<br\/> <b> B\u01b0\u1edbc 2:<\/b> Ph\u00e2n t\u00edch \u0111a th\u1ee9c theo c\u00e1ch \u0111\u1eb7t nh\u00e2n t\u1eed chung. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> $\\begin{align} & p^6q-p^5q^3-p^2q^4+pq^6 \\\\ & =(p^6q-p^5q^3)-(p^2q^4-pq^6) \\\\ & =p^5q(p-q^2)-pq^4(p-q^2) \\\\ & =pq(p-q^2)(p^4-q^3) \\\\ \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady ch\u1ecdn \u0111\u00e1p \u00e1n B. <\/span><\/span> ","column":2}]}],"id_ques":779},{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai10/lv3/img\/3.jpg' \/><\/center> K\u1ebft qu\u1ea3 khai tri\u1ec3n $(x^2+y^2)^2-(2xy)^2$ l\u00e0: ","select":["A. $(x+y)^2$ ","B. $x^4-y^4$","C. $(x^2+y^2)(x+y)(x-y)$","D. $(x+y)^2(x-y)^2$"],"hint":" Khai tri\u1ec3n theo h\u1eb1ng \u0111\u1eb3ng th\u1ee9c: $a^2-b^2=(a+b)(a-b)$","explain":"<span class='basic_left'> $\\begin{align} & (x^2+y^2)^2-(2xy)^2 \\\\ & = (x^2+y^2+2xy)(x^2+y^2-2xy) \\\\ & = (x+y)^2(x-y)^2\\\\ \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady ch\u1ecdn \u0111\u00e1p \u00e1n D. <\/span><\/span> ","column":2}]}],"id_ques":780}],"lesson":{"save":0,"level":3}}