{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank","correct":[[["4x-4","-4+4x","4(x-1)","4(-1+x)"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai9/lv3/img\/3.jpg' \/><\/center> Ph\u00e9p chia $\\left( 6x^3+4x^2-9x-9\\right):(2x^2-2x-1)$ c\u00f3 d\u01b0 l\u00e0 _input_","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> \u0110\u1eb7t ph\u00e9p chia v\u00e0 t\u00ecm d\u01b0.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> $\\left. \\begin{align} & \\begin{matrix} 6{{x}^{3}}+4{{x}^{2}}-9x-9\\,\\,\\,\\,\\,\\, \\\\ \\,\\,\\,6{{x}^{3}}-6{{x}^{2}}\\,\\,-3x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\\\\\end{matrix} \\\\ & \\overline{\\begin{align} & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,10{{x}^{2}}-6x-9\\,\\,\\,\\,\\,\\,\\,\\,\\, \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,10{{x}^{2}}-10x-5 \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\overline{\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,4x-4\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,} \\\\ \\end{align}} \\\\ \\end{align} \\right|\\begin{matrix} \\dfrac{2{{x}^{2}}-2x-1}{3x+5} \\\\ \\begin{matrix} {} \\\\ {} \\\\\\end{matrix} \\\\ {} \\\\\\end{matrix}$ <br\/> Ph\u00e9p chia c\u00f3 d\u01b0 l\u00e0 $4x-4$. <br\/> <span class='basic_pink'> Do \u0111\u00f3 d\u01b0 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $4x-4$. <\/span><\/span> "}]}],"id_ques":721},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank","correct":[[["2x+3","3+2x"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai9/lv3/img\/2.jpg' \/><\/center> Ph\u00e9p chia $\\left( 4x^3-10x^2+10x+1\\right):$$(2x^2-4x+2)$ c\u00f3 d\u01b0 l\u00e0 _input_","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> \u0110\u1eb7t ph\u00e9p chia v\u00e0 t\u00ecm d\u01b0.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> $\\left. \\begin{align} & \\begin{matrix} 4{{x}^{3}}-10{{x}^{2}}+10x+1 \\\\ \\,\\,\\,4{{x}^{3}}-8{{x}^{2}}\\,\\,+4x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\\\\\end{matrix} \\\\ & \\overline{\\begin{align} & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-2{{x}^{2}}+6x+1\\,\\,\\, \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-2{{x}^{2}}+4x-2 \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\overline{\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2x+3} \\\\ \\end{align}} \\\\ \\end{align} \\right|\\begin{matrix} \\dfrac{2{{x}^{2}}-4x+2}{2x-1} \\\\ \\begin{matrix} {} \\\\ {} \\\\\\end{matrix} \\\\ {} \\\\\\end{matrix}$<br\/> Ph\u00e9p chia c\u00f3 d\u01b0 l\u00e0 $2x+3$. <br\/> <span class='basic_pink'> Do \u0111\u00f3 d\u01b0 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $2x+3$. <\/span><\/span> "}]}],"id_ques":722},{"time":24,"part":[{"title":"\u0110i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank","correct":[[["20"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai9/lv3/img\/12.jpg' \/><\/center> \u0110\u1ec3 ph\u00e9p chia $(3x^2+mx+27):(x+5)$ c\u00f3 d\u01b0 l\u00e0 $2$ th\u00ec $m$ = _input_ ","hint":" Th\u1ef1c hi\u1ec7n ph\u00e9p chia $(3x^2+mx+27):(x+5)$, t\u00ecm d\u01b0. <br\/> Cho d\u01b0 b\u1eb1ng $2$ \u0111\u1ec3 t\u00ecm $m$.","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 $(3x^2+mx+27):(x+5)$, t\u00ecm d\u01b0.<br\/> <b> B\u01b0\u1edbc 2:<\/b> Cho s\u1ed1 d\u01b0 \u0111\u00f3 b\u1eb1ng 2 \u0111\u1ec3 t\u00ecm m. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> Ta th\u1ef1c hi\u1ec7n ph\u00e9p chia <br\/> $\\left. \\begin{align} & \\begin{matrix} 3{{x}^{2}}+mx+27 \\\\ 3{{x}^{2}}+15x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\\\\\end{matrix} \\\\ & \\overline{\\begin{align} & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( m-15 \\right)x+27 \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( m-15 \\right)x+5m-75 \\\\ &\\,\\,\\,\\,\\,\\,\\,\\,\\,\\overline{\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,102-5m\\,\\,\\,\\,\\,\\,\\,} \\\\ \\end{align}} \\\\ \\end{align} \\right|\\begin{matrix} \\dfrac{x+5}{3x+m-15} \\\\ \\begin{matrix} {} \\\\ {} \\\\\\end{matrix} \\\\ {} \\\\\\end{matrix}$. <br\/> V\u1eady $(3x^2+mx+27):(x+5)$$=3x+m-15$ d\u01b0 $102-5m$.<br\/> \u0110\u1ec3 $(3x^2+mx+27):(x+5)$ c\u00f3 d\u01b0 l\u00e0 $2$ th\u00ec $102-5m=2 \\Rightarrow m=20$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $20$. <\/span><\/span> "}]}],"id_ques":723},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank","correct":[[["6"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai9/lv3/img\/16.jpg' \/><\/center>Bi\u1ebft $(16z^2-16yz+4y^2)$$:(4z-2y)-4z+12=0$, gi\u00e1 tr\u1ecb $y$ 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> Th\u1ef1c hi\u1ec7n ph\u00e9p chia $(16z^2-16yz+4y^2):(4z-2y)$, \u0111\u01b0a $16z^2-16yz+4y^2$ v\u1ec1 b\u00ecnh ph\u01b0\u01a1ng c\u1ee7a m\u1ed9t hi\u1ec7u. <br\/> <b> B\u01b0\u1edbc 2:<\/b> R\u00fat g\u1ecdn \u0111\u1ec3 t\u00ecm gi\u00e1 tr\u1ecb c\u1ee7a $y$. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> Ta c\u00f3:<br\/> $ (16z^2-16yz+4y^2):(4z-2y)-4z+12=0 $<br\/>$\\Leftrightarrow(4z-2y)^2:(4z-2y)-4z+12=0 $<br\/>$\\Leftrightarrow 4z-2y-4z+12=0 $<br\/>$\\Leftrightarrow-2y+12=0 $<br\/>$ \\Leftrightarrow y=6 $ <br\/> <span class='basic_pink'> Do \u0111\u00f3 s\u1ed1 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $6$. <\/span><br\/><b>Nh\u1eadn x\u00e9t:<\/b> V\u1edbi m\u1ed9t s\u1ed1 ph\u00e9p chia \u0111a th\u1ee9c cho \u0111a th\u1ee9c: \u0110a th\u1ee9c b\u1ecb chia c\u00f3 th\u1ec3 ph\u00e2n t\u00edch th\u00e0nh nh\u00e2n t\u1eed ch\u1ee9a nh\u00e2n t\u1eed l\u00e0 \u0111a th\u1ee9c chia, th\u00ec ta c\u00f3 th\u1ec3 s\u1eed d\u1ee5ng t\u00ednh ch\u1ea5t $a.b:b=a$ \u0111\u1ec3 th\u1ef1c hi\u1ec7n ph\u00e9p chia.<\/span> "}]}],"id_ques":724},{"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/bai9/lv3/img\/11.jpg' \/><\/center> K\u1ebft qu\u1ea3 ph\u00e9p chia $(x^3-7x-6):(x+1)$ l\u00e0: ","select":["A. $x^2-7x+1$ ","B. $x^2-6x-1$ ","C. $x^2-2x-5$","D. $x^2-x-6$"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/> T\u00e1ch $x^3-7x-6=x^3+1-7x-7$, ph\u00e2n t\u00edch ti\u1ebfp th\u00e0nh nh\u00e2n t\u1eed b\u1eb1ng c\u00e1ch nh\u00f3m v\u00e0 \u0111\u1eb7t nh\u00e2n t\u1eed chung. <br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'> Ta c\u00f3:<br\/> $\\begin{align} & ({{x}^{3}}-7x-6):(x+1) \\\\ & =\\left[ {{x}^{3}}+1-7x-7 \\right]:\\left( x+1 \\right) \\\\ & =\\left[ \\left( x+1 \\right)\\left( {{x}^{2}}-x+1 \\right)-7\\left( x+1 \\right) \\right]:\\left( x+1 \\right) \\\\ & =\\left( x+1 \\right)\\left( {{x}^{2}}-x-6 \\right):\\left( x-1 \\right) \\\\ & ={{x}^{2}}-x-6 \\\\ \\end{align}$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span>","column":2}]}],"id_ques":725},{"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/bai9/lv3/img\/9.jpg' \/><\/center> K\u1ebft qu\u1ea3 ph\u00e9p chia $(9x^4-6x^3-3x^2+2x):(3x^2-1)$ l\u00e0: ","select":["A. $3x-2$ ","B. $x(3x-2)$ ","C. $x(3x+2)$","D. $3x+2$"],"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 \u0111a th\u1ee9c $9x^4-6x^3-3x^2+2x$ th\u00e0nh nh\u00e2n t\u1eed b\u1eb1ng c\u00e1ch nh\u00f3m v\u00e0 \u0111\u1eb7t nh\u00e2n t\u1eed chung.<br\/> <b> B\u01b0\u1edbc 2:<\/b> Th\u1ef1c hi\u1ec7n ph\u00e9p chia sau khi ph\u00e2n t\u00edch. <br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'>Ta c\u00f3 :<br\/> $ (9{{x}^{4}}-6{{x}^{3}}-3{{x}^{2}}+2x):(3{{x}^{2}}-1) $<br\/>$ =\\left[ 3{{x}^{3}}\\left( 3x-2 \\right)-x\\left( 3x-2 \\right) \\right]:\\left( 3{{x}^{2}}-1 \\right) $<br\/>$ =\\left[ x\\left( 3x-2 \\right)\\left( 3{{x}^{2}}-1 \\right) \\right]:\\left( 3{{x}^{2}}-1 \\right) $<br\/>$ =x\\left( 3x-2 \\right) $<span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span>","column":2}]}],"id_ques":726},{"time":24,"part":[{"title":"\u0110i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng \u0111\u1ec3 ho\u00e0n th\u00e0nh ph\u00e9p chia sau ","title_trans":"","temp":"fill_the_blank","correct":[[["x-1"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai9/lv3/img\/8.jpg' \/><\/center> $(4x^3-19x+15):(\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{})=4x^2+4x-15$","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 $4x^3-19x+15$ th\u00e0nh nh\u00e2n t\u1eed b\u1eb1ng c\u00e1ch t\u00e1ch $-19x=-4x-15x$ r\u1ed3i nh\u00f3m, \u0111\u1eb7t nh\u00e2n t\u1eed chung.<br\/> <b> B\u01b0\u1edbc 2:<\/b> Th\u1ef1c hi\u1ec7n ph\u00e9p chia $(4x^3-19x+15):(4x^2+4x-15)$. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> Ta c\u00f3: <br\/> $ (4x^3-19x+15):(4x^2+4x-15) \\\\ =(4x^3-4x-15x+15):(4x^2+4x-15) \\\\ =[4x(x^2-1)-15(x-1)]:(4x^2+4x-15) \\\\ =[4x(x-1)(x+1)-15(x-1)]:(4x^2+4x-15) \\\\ =[(x-1)(4x^2+4x-15)]:(4x^2+4x-15) \\\\ =x-1 $ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $x-1$. <\/span><\/span> "}]}],"id_ques":727},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"select":["A. $2x^2-2$","B. $x^2+2x$","C. $x^2-2x$"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai9/lv3/img\/5.jpg' \/><\/center> $(x^5-4x^3-5x^2+10x):$(?)$=x^3+2x^2-5$","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^5-4x^3-5x^2+10x$ th\u00e0nh nh\u00e2n t\u1eed b\u1eb1ng c\u00e1ch nh\u00f3m $(x^5-4x^3)-(5x^2-10x)$ r\u1ed3i \u0111\u1eb7t nh\u00e2n t\u1eed chung.<br\/> <b> B\u01b0\u1edbc 2:<\/b> Th\u1ef1c hi\u1ec7n ph\u00e9p chia $(x^5-4x^3-5x^2+10x):(x^3+2x^2-5)$. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> Ta th\u1ef1c hi\u1ec7n ph\u00e9p chia <br\/> $ ({{x}^{5}}-4{{x}^{3}}-5{{x}^{2}}+10x):({{x}^{3}}+2{{x}^{2}}-5) \\\\ =\\left[ {{x}^{3}}\\left( {{x}^{2}}-4 \\right)-5x\\left( x-2 \\right) \\right]:\\left( {{x}^{3}}+2{{x}^{2}}-5 \\right) \\\\ =x\\left( x-2 \\right)\\left[ {{x}^{2}}\\left( x+2 \\right)-5 \\right]:\\left( {{x}^{3}}+2{{x}^{2}}-5 \\right) \\\\ =\\left[ x\\left( x-2 \\right)\\left( {{x}^{3}}+2{{x}^{2}}-5 \\right) \\right]:\\left( {{x}^{3}}+2{{x}^{2}}-5 \\right) \\\\ =x\\left( x-2 \\right) \\\\ ={{x}^{2}}-2x $<\/span> "}]}],"id_ques":728},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank","correct":[[["3x-1"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai9/lv3/img\/4.jpg' \/><\/center> K\u1ebft qu\u1ea3 ph\u00e9p chia <br\/> $(6x^2+13x-5):(2x+5)=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1:<\/b> T\u00e1ch $6x^2+13x-5$$=6x^2-2x+15x-5$, ph\u00e2n t\u00edch ti\u1ebfp th\u00e0nh nh\u00e2n t\u1eed b\u1eb1ng c\u00e1ch nh\u00f3m r\u1ed3i \u0111\u1eb7t nh\u00e2n t\u1eed chung.<br\/> <b> B\u01b0\u1edbc 2:<\/b> Th\u1ef1c hi\u1ec7n ph\u00e9p chia.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> $\\begin{align} & (6{{x}^{2}}+13x-5):(2x+5) \\\\ & =\\left( 6{{x}^{2}}-2x+15x-5 \\right):\\left( 2x+5 \\right) \\\\ & =\\left[ 2x\\left( 3x-1 \\right)+5\\left( 3x-1 \\right) \\right]:\\left( 2x+5 \\right) \\\\ & =\\left( 3x-1 \\right)\\left( 2x+5 \\right):\\left( 2x+5 \\right) \\\\ & =3x-1 \\\\ \\end{align}$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 th\u01b0\u01a1ng ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $3x-1$. <\/span><\/span> "}]}],"id_ques":729},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $5a\u22125b+2$","B. $5a\u2212b+2$","C. $5a +b + 2$"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai9/lv3/img\/3.jpg' \/><\/center> K\u1ebft qu\u1ea3 ph\u00e9p chia <br\/> $[5(a-b)^3+2(a-b)^2]:(b-a)^2= ?$","hint":" Ph\u00e2n t\u00edch $5(a-b)^3+2(a-b)^2$ th\u00e0nh nh\u00e2n t\u1eed b\u1eb1ng c\u00e1ch \u0111\u1eb7t $(a-b)^2$ l\u00e0m nh\u00e2n t\u1eed chung.<br\/> Th\u1ef1c hi\u1ec7n ph\u00e9p chia.","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 $5(a-b)^3+2(a-b)^2$ th\u00e0nh nh\u00e2n t\u1eed b\u1eb1ng c\u00e1ch \u0111\u1eb7t $(a-b)^2$ l\u00e0m nh\u00e2n t\u1eed chung.<br\/> <b> B\u01b0\u1edbc 2:<\/b> Th\u1ef1c hi\u1ec7n ph\u00e9p chia.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> $\\begin{align} & [5{{(a-b)}^{3}}+2{{(a-b)}^{2}}]:{{(b-a)}^{2}} \\\\ & ={{\\left( a-b \\right)}^{2}}\\left[ 5\\left( a-b \\right)+2 \\right]:{{\\left( a-b \\right)}^{2}} \\\\ & =5\\left( a-b \\right)+2 \\\\ & =5a-5b+2 \\\\ \\end{align}$<\/span> "}]}],"id_ques":730}],"lesson":{"save":0,"level":3}}