{"segment":[{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai10/lv2/img\/5.jpg' \/><\/center> \u0110\u1ec3 $[(n^2+n+1)(n-4)+3]\\,\\vdots \\, (n^2+n+1)$ th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a $n$ l\u00e0: ","select":["A. $n=0;n=1;n=2$ ho\u1eb7c $n=3$ ","B. $n=\\pm 2$ ho\u1eb7c $n=\\pm 3$","C. $n=\\pm 1;n=-2$ ho\u1eb7c $n=0$","D. $n=\\pm 1$ ho\u1eb7c $n=\\pm 2$"],"hint":"","explain":"<span class='basic_left'> Ta c\u00f3: $[(n^2+n+1)(n-4)]\\,\\vdots \\, (n^2+n+1)$ <br\/> N\u00ean \u0111\u1ec3 $[(n^2+n+1)(n-4)+3]\\,\\vdots \\, (n^2+n+1)$ th\u00ec $3\\,\\vdots \\, (n^2+n+1)$ <br\/> M\u00e0 $n^2+n+1=(n+\\dfrac{1}{2})^2+\\dfrac{3}{4} > 0$ v\u1edbi m\u1ecdi $n$ <br\/> $\\Rightarrow$ $ \\left[ \\begin{aligned} & {{n}^{2}}+n+1=3 \\\\ & {{n}^{2}}+n+1=1 \\\\ \\end{aligned} \\right.$ $\\Leftrightarrow \\left[ \\begin{aligned} & {{n}^{2}}+n-2=0 \\\\ & {{n}^{2}}+n=0 \\\\ \\end{aligned} \\right.$ $\\Leftrightarrow \\left[ \\begin{aligned} & \\left( n-1 \\right)\\left( n+2 \\right)=0 \\\\ & n\\left( n+1 \\right)=0 \\\\ \\end{aligned} \\right.$ $\\Leftrightarrow \\left[ \\begin{aligned} & n=1 \\\\ & n=-2 \\\\ & n=0 \\\\ & n=-1 \\\\ \\end{aligned} \\right.$ <br\/> <span class='basic_pink'> V\u1eady ch\u1ecdn \u0111\u00e1p \u00e1n C. <\/span><\/span> ","column":2}]}],"id_ques":751},{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai10/lv2/img\/5.jpg' \/><\/center> \u0110\u1ec3 $(2x^2+x-7)\\,\\vdots \\, (x-2)$ th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a $x$ l\u00e0: ","select":["A. $x=-1;x=1;x=3$ ho\u1eb7c $x =5$ ","B. $x=1;x=3;x=5$ ho\u1eb7c $x =7$","C. $x=-1;x=1$ ho\u1eb7c $x =0$","D. $x=\\pm 1;x=\\pm 2$ ho\u1eb7c $x= \\pm 3$"],"hint":"","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 $(2x^2+x-7)\\,: \\, (x-2)$, t\u00ecm s\u1ed1 d\u01b0 <br\/> <b> B\u01b0\u1edbc 2:<\/b> Cho $x-2$ l\u00e0 \u01b0\u1edbc c\u1ee7a s\u1ed1 d\u01b0 \u0111\u1ec3 t\u00ecm $x$. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> $\\left. \\begin{align} & \\begin{matrix} 2{{x}^{2}}+x-7\\,\\,\\,\\,\\,\\,\\,\\,\\, \\\\ 2{{x}^{2}}-4x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\\\\\end{matrix} \\\\ & \\overline{\\begin{align} & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,5x-7\\,\\,\\,\\,\\, \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,5x-10\\,\\, \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\overline{\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,3\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,} \\\\ \\end{align}} \\\\ \\end{align} \\right|\\begin{matrix} \\dfrac{x-2}{2x+5} \\\\ \\begin{matrix} {} \\\\ {} \\\\\\end{matrix} \\\\ {} \\\\\\end{matrix}$<br\/> \u0110\u1ec3 $(2x^2+x-7)\\,\\vdots \\, (x-2)$ th\u00ec $3\\,\\vdots (x-2)$ hay $(x-2)\\in \u01af(3)=\\left\\{ \\text{-1;1;-3;3} \\right\\} $<br\/> $\\Rightarrow$ $\\left[ \\begin{aligned} & x-2=-1 \\\\ & x-2=1 \\\\ & x-2=-3 \\\\ & x-2=3 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left[ \\begin{aligned} & x=1 \\\\ & x=3 \\\\ & x=-1 \\\\ & x=5 \\\\ \\end{aligned} \\right.$ <br\/> <span class='basic_pink'> V\u1eady ch\u1ecdn \u0111\u00e1p \u00e1n A. <\/span><\/span> ","column":2}]}],"id_ques":752},{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai10/lv2/img\/4.jpg' \/><\/center> T\u00ecm $m, n$ $( m, n\\,\\in\\,\\mathbb{N})$ \u0111\u1ec3 ph\u00e9p chia sau l\u00e0 ph\u00e9p chia h\u1ebft:<br\/> $(x^ny^{m+1}+x^3y^5+x^6y^6):x^2y^2$ ","select":["A. $m, n\\,\\in\\,\\mathbb{N};\\,m\\,\\ge 2;n \\ge 1 $","B. $m, n\\,\\in\\,\\mathbb{N};\\,m\\,\\ge 1;n\\, \\ge \\,2 $","C. $m, n\\,\\in\\,\\mathbb{N};\\,m\\,\\le 1; n \\ge 2 $","D. $m, n\\,\\in\\,\\mathbb{N};\\,m\\,\\le 1;n \\ge 3 $"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/> <b> Quy t\u1eafc:<\/b> \u0110\u1ec3 $ax^my^n$ chia h\u1ebft cho $bx^py^q$ th\u00ec $\\left\\{ \\begin{align} & m\\ge p \\\\ & n\\ge q \\\\ \\end{align} \\right.$ <br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'>\u0110\u1ec3 $(x^ny^{m+1}+x^3y^5+x^6y^6):x^2y^2$ l\u00e0 ph\u00e9p chia h\u1ebft th\u00ec <br\/>$\\left\\{ \\begin{aligned} & {{x}^{n}}{{y}^{m+1}}\\,\\,\\,\\vdots \\,\\,{{x}^{2}}{{y}^{2}} \\\\ & {{x}^{3}}{{y}^{5}}\\,\\,\\,\\vdots \\,\\,{{x}^{2}}{{y}^{2}} \\\\ & {{x}^{6}}{{y}^{6}}\\,\\,\\vdots \\,\\,{{x}^{2}}{{y}^{2}} \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & n\\ge 2 \\\\ & m+1\\ge 2 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m\\ge 1 \\\\ & n\\ge 2 \\\\ \\end{aligned} \\right.$ v\u1edbi $m, n\\in \\mathbb{N}$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span>","column":2}]}],"id_ques":753},{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai10/lv2/img\/4.jpg' \/><\/center> T\u00ecm $m, n$ $( m, n\\,\\in\\,\\mathbb{N})$ \u0111\u1ec3 ph\u00e9p chia sau l\u00e0 ph\u00e9p chia h\u1ebft:<br\/> $\\dfrac{1}{2}x^{m+1}y^nz^2:3x^2y^3$ ","select":["A. $m, n\\,\\in\\,\\mathbb{N};\\,m\\,\\ge 1,\\,n \\ge 3 $","B. $m, n\\,\\in\\,\\mathbb{N};\\,m\\,>\\,1,\\,n\\, > \\,3 $","C. $m, n\\,\\in\\,\\mathbb{N};\\,m\\,\\le 1,\\,n \\le 3 $","D. $m, n\\,\\in\\,\\mathbb{N};\\,m\\,< 1,\\,n < 3 $"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/> <b> Quy t\u1eafc:<\/b> \u0110\u1ec3 $ax^my^n$ chia h\u1ebft cho $bx^py^q$ th\u00ec $\\left\\{ \\begin{align} & m\\ge p \\\\ & n\\ge q \\\\ \\end{align} \\right.$ <br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'>\u0110\u1ec3 $\\dfrac{1}{2}x^{m+1}y^nz^2:3x^2y^3$ l\u00e0 ph\u00e9p chia h\u1ebft th\u00ec <br\/>$\\left\\{ \\begin{aligned} & m+1\\ge 2 \\\\ & n\\ge 3 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m\\ge 1 \\\\ & n\\ge 3 \\\\ \\end{aligned} \\right.$ v\u1edbi $m, n\\in \\mathbb{N}$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span>","column":2}]}],"id_ques":754},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $6u^3+8v\u221218uv$","B. $6u^2+8v\u221218uv$","C. $6u^3-8v\u221218uv$"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai10/lv2/img\/3.jpg' \/><\/center> $\\left( 3{{u}^{5}}{{v}^{2}}+4{{u}^{2}}{{v}^{3}}-9{{u}^{3}}{{v}^{3}} \\right):\\dfrac{1}{2}{{u}^{2}}{{v}^{2}}=?$","hint":" Th\u1ef1c hi\u1ec7n ph\u00e9p chia \u0111a th\u1ee9c cho \u0111\u01a1n th\u1ee9c (trong tr\u01b0\u1eddng h\u1ee3p c\u00e1c h\u1ea1ng t\u1eed c\u1ee7a \u0111a th\u1ee9c \u0111\u1ec1u chia h\u1ebft cho \u0111\u01a1n th\u1ee9c):<br\/> <b> $(A+ B+C):D=A:D+B:D+C:D$ <\/b>","explain":"<span class='basic_left'><br\/> $\\begin{align} & \\left( 3{{u}^{5}}{{v}^{2}}+4{{u}^{2}}{{v}^{3}}-9{{u}^{3}}{{v}^{3}} \\right):\\dfrac{1}{2}{{u}^{2}}{{v}^{2}} \\\\ & =3{{u}^{5}}{{v}^{2}}:\\dfrac{1}{2}{{u}^{2}}{{v}^{2}}+4{{u}^{2}}{{v}^{3}}:\\dfrac{1}{2}{{u}^{2}}{{v}^{2}}-9{{u}^{3}}{{v}^{3}}:\\dfrac{1}{2}{{u}^{2}}{{v}^{2}} \\\\ & =6{{u}^{3}}+8v-18uv \\\\ \\end{align}$<\/span> "}]}],"id_ques":755},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"select":["A. $\\dfrac{1}{5}xy+\\dfrac{2}{5}y+\\dfrac{7}{10}x^2$","B. $\\dfrac{1}{5}xy+\\dfrac{1}{15}y+\\dfrac{7}{10}x^2$","C. $\\dfrac{2}{5}xy+\\dfrac{1}{15}y+\\dfrac{3}{10}x^2$"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai10/lv2/img\/2.jpg' \/><\/center> $\\left( {{x}^{2}}{{y}^{2}}+\\dfrac{1}{3}x{{y}^{2}}+\\dfrac{7}{2}{{x}^{3}}y \\right):5xy=?$","hint":" Th\u1ef1c hi\u1ec7n ph\u00e9p chia \u0111a th\u1ee9c cho \u0111\u01a1n th\u1ee9c (trong tr\u01b0\u1eddng h\u1ee3p c\u00e1c h\u1ea1ng t\u1eed c\u1ee7a \u0111a th\u1ee9c \u0111\u1ec1u chia h\u1ebft cho \u0111\u01a1n th\u1ee9c):<br\/> <b> $(A+ B+C):D=A:D+B:D+C:D$ <\/b>","explain":"<span class='basic_left'><br\/> $\\begin{align} & \\left( {{x}^{2}}{{y}^{2}}+\\dfrac{1}{3}x{{y}^{2}}+\\dfrac{7}{2}{{x}^{3}}y \\right):5xy \\\\ & ={{x}^{2}}{{y}^{2}}:5xy+\\dfrac{1}{3}x{{y}^{2}}:5xy+\\dfrac{7}{2}{{x}^{3}}y:5xy \\\\ & =\\dfrac{1}{5}xy+\\dfrac{1}{15}y+\\dfrac{7}{10}{{x}^{2}} \\\\ \\end{align}$ <\/span> "}]}],"id_ques":756},{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai10/lv2/img\/16.jpg' \/><\/center> K\u1ebft qu\u1ea3 ph\u00e2n t\u00edch \u0111a th\u1ee9c $(a+b)^3-(a-b)^3$ th\u00e0nh nh\u00e2n t\u1eed l\u00e0: ","select":["A. $(a+b)(a-b)$ ","B. $2a(a^2+3b^2)$","C. $2b(3a^2+b^2)$","D. $(a-b)(a^2+b^2)$"],"hint":"","explain":"<span class='basic_left'><br\/> $\\begin{align} & {{(a+b)}^{3}}-{{(a-b)}^{3}} \\\\ & =\\left( a+b-a+b \\right)\\left[ {{\\left( a+b \\right)}^{2}}+\\left( a+b \\right)\\left( a-b \\right)+{{\\left( a-b \\right)}^{2}} \\right] \\\\ & =2b\\left( {{a}^{2}}+2ab+{{b}^{2}}+{{a}^{2}}-{{b}^{2}}+{{a}^{2}}-2ab+{{b}^{2}} \\right) \\\\ & =2b\\left( 3{{a}^{2}}+{{b}^{2}} \\right) \\\\ \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady ch\u1ecdn \u0111\u00e1p \u00e1n C. <\/span><\/span> ","column":2}]}],"id_ques":757},{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai10/lv2/img\/13.jpg' \/><\/center> K\u1ebft qu\u1ea3 ph\u00e2n t\u00edch \u0111a th\u1ee9c $-x^3+9x^2-27x+27$ th\u00e0nh nh\u00e2n t\u1eed l\u00e0: ","select":["A. $x^3-3^3$ ","B. $27-x^3$","C. $(x-3)^3$","D. $(3-x)^3$"],"hint":" Ph\u00e2n t\u00edch $-x^3+9x^2-27x+27$ theo c\u00e1ch d\u00f9ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c. ","explain":"<span class='basic_left'><br\/> $\\begin{align} & -x^3+9x^2-27x+27 \\\\ & =27-27x+9x^2-x^3 \\\\ & =(3-x)^3 \\\\ \\end{align}$. <br\/> <span class='basic_pink'> V\u1eady ch\u1ecdn \u0111\u00e1p \u00e1n D. <\/span><\/span> ","column":2}]}],"id_ques":758},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t.<br\/> Kh\u1eb3ng \u0111\u1ecbnh d\u01b0\u1edbi \u0111\u00e2y \u0111\u00fang hay sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai10/lv2/img\/12.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $9y^2+6y+3$ l\u00e0 m\u1ed9t s\u1ed1 d\u01b0\u01a1ng v\u1edbi m\u1ecdi $y$ ","select":["\u0110\u00fang","Sai"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/> \u0110\u01b0a $9y^2+6y+3$ \u0111\u01b0a v\u1ec1 d\u1ea1ng $[f(y)]^2+a$ .<br\/> Nh\u1eadn x\u00e9t, \u0111\u00e1nh gi\u00e1 gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c. <br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'>Ta c\u00f3:<br\/> $\\begin{align} & 9y^2+6y+3 \\\\ & =(3y)^2+2.3y.1+1^2+2 \\\\ & =(3y+1)^2+2 \\\\ & \\text{Do} \\,\\,(3y+1)^2\\,\\,\\ge \\,\\,0\\,\\,\\Rightarrow (3y+1)^2+2\\,\\,\\,\\ge \\,\\,2 \\\\ \\end{align}$ <br\/> V\u1eady gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c l\u00e0 m\u1ed9t s\u1ed1 d\u01b0\u01a1ng v\u1edbi m\u1ecdi $y$. <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang.<\/span>","column":2}]}],"id_ques":759},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t.<br\/> Kh\u1eb3ng \u0111\u1ecbnh d\u01b0\u1edbi \u0111\u00e2y \u0111\u00fang hay sai","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai10/lv2/img\/12.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $\\dfrac{1}{9}a^2+\\dfrac{1}{6}ab^5+\\dfrac{1}{16}b^{10}$ l\u00e0 m\u1ed9t s\u1ed1 d\u01b0\u01a1ng v\u1edbi m\u1ecdi $a, b$ ","select":["\u0110\u00fang","Sai"],"hint":" Ph\u00e2n t\u00edch bi\u1ec3u th\u1ee9c theo h\u1eb1ng \u0111\u1eb3ng th\u1ee9c b\u00ecnh ph\u01b0\u01a1ng c\u1ee7a m\u1ed9t t\u1ed5ng r\u1ed3i \u0111\u00e1nh gi\u00e1.","explain":"<span class='basic_left'>Ta c\u00f3:<br\/> $\\begin{align} & \\dfrac{1}{9}{{a}^{2}}+\\dfrac{1}{6}a{{b}^{5}}+\\dfrac{1}{16}{{b}^{10}} \\\\ & ={{\\left( \\dfrac{1}{3}a \\right)}^{2}}+2.\\dfrac{1}{3}a.\\dfrac{1}{4}{{b}^{5}}+{{\\left( \\dfrac{1}{4}{{b}^{5}} \\right)}^{2}} \\\\ & ={{\\left( \\dfrac{1}{3}a+\\dfrac{1}{4}{{b}^{5}} \\right)}^{2}} \\\\ & \\text {Do} \\,\\,{{\\left( \\dfrac{1}{3}a+\\dfrac{1}{4}{{b}^{5}} \\right)}^{2}}\\,\\,\\ge \\,\\,0\\,\\,\\Rightarrow \\dfrac{1}{9}{{a}^{2}}+\\dfrac{1}{6}a{{b}^{5}}+\\dfrac{1}{16}{{b}^{10}}\\,\\,\\,\\ge \\,\\,0 \\\\ \\end{align}$ <br\/> V\u1eady gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c l\u00e0 m\u1ed9t s\u1ed1 kh\u00f4ng \u00e2m v\u1edbi m\u1ecdi $a, b$. <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 Sai.<\/span>","column":2}]}],"id_ques":760},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t.<br\/> Kh\u1eb3ng \u0111\u1ecbnh d\u01b0\u1edbi \u0111\u00e2y \u0111\u00fang hay sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai10/lv2/img\/11.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $16x^4-40x^2y^3+25y^6$ l\u00e0 m\u1ed9t s\u1ed1 kh\u00f4ng \u00e2m v\u1edbi m\u1ecdi $x, y$ ","select":["\u0110\u00fang","Sai"],"hint":" Ph\u00e2n t\u00edch $16x^4-40x^2y^3+25y^6$ \u0111\u01b0a v\u1ec1 b\u00ecnh ph\u01b0\u01a1ng c\u1ee7a m\u1ed9t hi\u1ec7u. ","explain":"<span class='basic_left'>Ta c\u00f3:<br\/> $\\begin{align} & 16{{x}^{4}}-40{{x}^{2}}{{y}^{3}}+25{{y}^{6}} \\\\ & ={{\\left( 4{{x}^{2}} \\right)}^{2}}-2.4{{x}^{2}}.5{{y}^{3}}+{{\\left( 5{{y}^{3}} \\right)}^{2}} \\\\ & ={{\\left( 4{{x}^{2}}-5{{y}^{3}} \\right)}^{2}} \\\\ \\end{align}$. <br\/> Do $(4x^2-5y^3)^2\\ge 0$ n\u00ean gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c l\u00e0 m\u1ed9t s\u1ed1 kh\u00f4ng \u00e2m v\u1edbi m\u1ecdi $x, y$. <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang<\/span>","column":2}]}],"id_ques":761},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\left[\\begin{array}{l}{x=-5} \\\\ {x = \\dfrac{-23}{7}}\\end{array}\\right.$","B. $\\left[\\begin{array}{l}{x= 2} \\\\ {x = \\dfrac{-1}{2}}\\end{array}\\right.$","C. $\\left[\\begin{array}{l}{x=5} \\\\ {x = \\dfrac{-1}{2}}\\end{array}\\right.$"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai10/lv2/img\/10.jpg' \/><\/center> T\u00ecm $x$, bi\u1ebft $4(2x+7)^2-9(x+3)^2=0$ ","hint":"Bi\u1ebfn \u0111\u1ed5i v\u1ebf tr\u00e1i v\u1ec1 d\u1ea1ng A.B = 0 \u0111\u1ec3 t\u00ecm $x$","explain":"<span class='basic_left'><br\/> Ta c\u00f3: <br\/> $\\begin{align} & 4(2x+7)^2-9(x+3)^2=0 \\\\ &\\Leftrightarrow 4(4x^2+28x+49)-9(x^2+6x+9)=0 \\\\ &\\Leftrightarrow 16x^2+112x+196-9x^2-54x-81=0 \\\\ &\\Leftrightarrow 7x^2+58x+115=0 \\\\ &\\Leftrightarrow 7x^2+35x+23x+115=0 \\\\ &\\Leftrightarrow 7x(x+5)+23(x+5)=0 \\\\ &\\Leftrightarrow (x+5)(7x+23)=0 \\\\ \\Leftrightarrow \\left[ \\begin{aligned} & x+5=0 \\\\ & 7x+23=0 \\\\ \\end{aligned} \\right. \\Leftrightarrow \\left[ \\begin{aligned} & x=-5 \\\\ & x=-\\dfrac{23}{7} \\\\ \\end{aligned} \\right. \\\\ \\end{align}$<\/span> "}]}],"id_ques":762},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank","correct":[[["7"],["4"]]],"list":[{"point":5,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai10/lv2/img\/9.jpg' \/><\/center> T\u00ecm $y$, bi\u1ebft $(3y^2-y+1)(y-1)+y^2(4-3y)=\\dfrac{5}{2}$ <br\/> \u0110\u00e1p \u00e1n: $y$ = <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span> <br\/> Bi\u1ebfn \u0111\u1ed5i, r\u00fat g\u1ecdn \u0111\u01b0a v\u1ec1 d\u1ea1ng $ay+b=0$ \u0111\u1ec3 t\u00ecm $y$. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> Ta c\u00f3: <br\/> $\\begin{align} (3{{y}^{2}}-y+1)(y-1)+{{y}^{2}}(4-3y) &=\\dfrac{5}{2} \\\\ 3{{y}^{3}}-3{{y}^{2}}-{{y}^{2}}+y+y-1+4{{y}^{2}}-3{{y}^{3}}&=\\dfrac{5}{2} \\\\ 2y-1&=\\dfrac{5}{2} \\\\ 2y & =\\dfrac{7}{2} \\\\ y&=\\dfrac{7}{4} \\\\ \\end{align}$ <\/span> "}]}],"id_ques":763},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank_random","correct":[[["0"],["2"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai10/lv2/img\/9.jpg' \/><\/center> T\u00ecm $x$, bi\u1ebft $x^2-4+(x-2)^2=0$ <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":"","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 $x^2-4+(x-2)^2$ th\u00e0nh nh\u00e2n t\u1eed b\u1eb1ng c\u00e1ch d\u00f9ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c v\u00e0 \u0111\u1eb7t nh\u00e2n t\u1eed chung.<br\/> <b> B\u01b0\u1edbc 2:<\/b> N\u1ebfu $a.b=0$ th\u00ec $\\left[ \\begin{align} & a=0 \\\\ & b=0 \\\\ \\end{align} \\right.$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> Ta c\u00f3: <br\/> $\\begin{aligned} & x^2-4+(x-2)^2=0 \\\\ & (x+2)(x-2)+(x-2)^2=0 \\\\ & (x-2)(x+2+x-2)=0 \\\\ & 2x(x-2)=0 \\\\ & \\Rightarrow \\left[ \\begin{aligned} & 2x=0 \\\\ & x-2=0 \\\\ \\end{aligned} \\right. \\Leftrightarrow \\left[ \\begin{aligned} & x=0 \\\\ & x=2 \\\\ \\end{aligned} \\right.\\end{aligned}$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $0$ v\u00e0 $2$. <\/span><\/span> "}]}],"id_ques":764},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank","correct":[[["96"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai10/lv2/img\/8.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $x^2y^2z+xy^2z^2+x^2yz^2$ t\u1ea1i $x=y=z=2$ l\u00e0 _input_","hint":"","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 $x^2y^2z+xy^2z^2+x^2yz^2$ th\u00e0nh nh\u00e2n t\u1eed b\u1eb1ng c\u00e1ch \u0111\u1eb7t nh\u00e2n t\u1eed chung.<br\/> <b> B\u01b0\u1edbc 2:<\/b> Thay $x=y=z=2$ v\u00e0o \u0111\u1ec3 t\u00ednh. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> Ta c\u00f3: <br\/> $\\begin{align} & x^2y^2z+xy^2z^2+x^2yz^2\\\\ &=xyz(xy+yz+xz) \\\\ \\end{align}$ <br\/> Thay $x=y=z=2$ v\u00e0o bi\u1ec3u th\u1ee9c $xyz(xy+yz+xz)$, ta \u0111\u01b0\u1ee3c:<br\/> $2.2.2(2.2+2.2+2.2)=8.12=96$. <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $96$. <\/span><\/span> "}]}],"id_ques":765},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank","correct":[[["416"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai10/lv2/img\/5.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 $9(x+5)^2-(x-7)^2$ t\u1ea1i $x=2$<br\/>Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c l\u00e0: _input_","hint":"","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 $9(x+5)^2-(x-7)^2$ th\u00e0nh nh\u00e2n t\u1eed b\u1eb1ng c\u00e1ch d\u00f9ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c.<br\/> <b>B\u01b0\u1edbc 2:<\/b> Thay $x=2$ v\u00e0o \u0111\u1ec3 t\u00ednh. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> Ta c\u00f3: <br\/> $\\begin{align} & 9{{(x+5)}^{2}}-{{(x-7)}^{2}}\\\\ &={{[3(x+5)]}^{2}}-{{(x-7)}^{2}} \\\\ & =(3x+15+x-7)(3x+15-x+7) \\\\ & =(4x+8)(2x+22) \\\\ & =4(x+2)2(x+11) \\\\ & =8(x+2)(x+11) \\\\ \\end{align}$ <br\/> Thay $x=2$ v\u00e0o bi\u1ec3u th\u1ee9c $8(x+2)(x+11)$, ta \u0111\u01b0\u1ee3c:<br\/> $8(2+2)(2+11)=8.4.13=416$. <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $416$. <\/span><\/span> "}]}],"id_ques":766},{"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":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai10/lv2/img\/5.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $9x^2+6x-8$ t\u1ea1i $x=\\dfrac{2}{3}$ l\u00e0 _input_","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span> <br\/> <b> C\u00e1ch 1: <\/b> Thay $x=\\dfrac{2}{3}$ v\u00e0o bi\u1ec3u th\u1ee9c r\u1ed3i t\u00ednh.<br\/> <b> C\u00e1ch 2:<\/b> <br\/> <b> B\u01b0\u1edbc 1:<\/b> Ph\u00e2n t\u00edch $9x^2+6x-8$ th\u00e0nh nh\u00e2n t\u1eed b\u1eb1ng c\u00e1ch t\u00e1ch $6x=-6x+12x$, r\u1ed3i \u0111\u1eb7t nh\u00e2n t\u1eed chung.<br\/> <b> B\u01b0\u1edbc 2:<\/b> Thay $x=\\dfrac{2}{3}$ v\u00e0o \u0111\u1ec3 t\u00ednh. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> <b> C\u00e1ch 1: <\/b> Thay $x=\\dfrac{2}{3}$ v\u00e0o bi\u1ec3u th\u1ee9c:<br\/> $9.(\\dfrac{2}{3})^2+6.\\dfrac{2}{3}-8=0$<br\/> <b> C\u00e1ch 2:<\/b> Ta c\u00f3: <br\/> $9x^2+6x-8=9x^2-6x+12x-8$$=3x(3x-2)+4(3x-2)=(3x-2)(3x+4)$<br\/> Thay $x=\\dfrac{2}{3}$ v\u00e0o bi\u1ec3u th\u1ee9c $(3x-2)(3x+4)$, ta \u0111\u01b0\u1ee3c:<br\/> $\\left(3.\\dfrac{2}{3}-2\\right) \\left(3.\\dfrac{2}{3}+4\\right)=0$. <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $0$. <\/span><\/span> "}]}],"id_ques":767},{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai10/lv2/img\/4.jpg' \/><\/center> K\u1ebft qu\u1ea3 ph\u00e2n t\u00edch \u0111a th\u1ee9c $4x^2-4xy+y^2-9$ th\u00e0nh nh\u00e2n t\u1eed l\u00e0 ","select":["A. $(2x-y+3)(2x-y-3)$ ","B. $(2x-y)(2x+y)$","C. $(2x+3)(2x-3)$","D. $(2x+y+3)(2x+y-3)$"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1:<\/b> Nh\u00f3m $4x^2-4xy+y^2-9=(4x^2-4xy+y^2)-9$.<br\/> <b> B\u01b0\u1edbc 2:<\/b> Ph\u00e2n t\u00edch \u0111a th\u1ee9c theo c\u00e1ch d\u00f9ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> $\\begin{align} & 4x^2-4xy+y^2-9 \\\\ & =(4x^2-4xy+y^2)-9 \\\\ & =(2x-y)^2-3^2 \\\\ & =(2x-y+3)(2x-y-3) \\\\ \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady ch\u1ecdn \u0111\u00e1p \u00e1n A. <\/span><\/span> ","column":2}]}],"id_ques":768},{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai10/lv2/img\/3.jpg' \/><\/center> K\u1ebft qu\u1ea3 khai tri\u1ec3n $(2x-\\dfrac{1}{2})^2$ l\u00e0 ","select":["A. $4x^2-x+\\dfrac{1}{2}$ ","B. $4x^2-2x+\\dfrac{1}{4}$","C. $x^2+4xy+y^2$","D. $4x^2+2x+\\dfrac{1}{2}$"],"hint":" Khai tri\u1ec3n theo h\u1eb1ng \u0111\u1eb3ng th\u1ee9c: $(a-b)^2=a^2-2ab+b^2$","explain":"<span class='basic_left'> $\\begin{align} & (2x-\\dfrac{1}{2})^2 \\\\ & = (2x)^2-2.2x.\\dfrac{1}{2}+(\\dfrac{1}{2})^2\\\\ & = 4x^2-2x+\\dfrac{1}{4}\\\\ \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady ch\u1ecdn \u0111\u00e1p \u00e1n B. <\/span><\/span> ","column":2}]}],"id_ques":769},{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai10/lv2/img\/2.jpg' \/><\/center> K\u1ebft qu\u1ea3 ph\u00e9p nh\u00e2n $(x^2+2xy-3)(-xy)$ l\u00e0 ","select":["A. $x^2y-2x^2y^2+3xy$ ","B. $-x^3y-2x^2y^2+3xy$","C. $x^3y-2x^2y^2-3xy$","D. $x^2y-2xy^2+3xy$"],"hint":" Nh\u00e2n \u0111a th\u1ee9c v\u1edbi \u0111\u01a1n th\u1ee9c: ta nh\u00e2n t\u1eebng h\u1ea1ng t\u1eed c\u1ee7a \u0111a th\u1ee9c \u0111\u00f3 v\u1edbi \u0111\u01a1n th\u1ee9c r\u1ed3i c\u1ed9ng c\u00e1c t\u00edch v\u1edbi nhau.","explain":"<span class='basic_left'><br\/> $\\begin{align} & (x^2+2xy-3)(-xy) \\\\ & =-x^3y-2x^2y^2+3xy \\\\ \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady ch\u1ecdn \u0111\u00e1p \u00e1n B. <\/span><\/span> ","column":2}]}],"id_ques":770}],"lesson":{"save":0,"level":2}}