{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-5"],["2"]]],"list":[{"point":10,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai13/lv3/img\/12.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $\\dfrac{3}{2{{x}^{2}}+6x}-\\dfrac{4-3{{x}^{2}}}{{{x}^{2}}-9}-3$ t\u1ea1i $x=1$ l\u00e0: <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b> B\u01b0\u1edbc 1: <\/b> T\u00ecm \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 bi\u1ec3u th\u1ee9c c\u00f3 ngh\u0129a.<br\/> <b> B\u01b0\u1edbc 2:<\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c sau \u0111\u00f3 thay $x=1$ v\u00e0o bi\u1ec3u th\u1ee9c \u0111\u00e3 r\u00fat g\u1ecdn \u0111\u1ec3 t\u00ednh. <br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x \\ne \\{-3;0;3\\}$ <br\/> Ta c\u00f3: <br\/> $ \\dfrac{3}{2{{x}^{2}}+6x}-\\dfrac{4-3{{x}^{2}}}{{{x}^{2}}-9}-3$$=\\dfrac{3}{2x\\left( x+3 \\right)}-\\dfrac{4-3{{x}^{2}}}{\\left( x+3 \\right)\\left( x-3 \\right)}-3 $<br\/>$ =\\dfrac{3\\left( x-3 \\right)}{2x\\left( x+3 \\right)\\left( x-3 \\right)}$$-\\dfrac{\\left( 4-3{{x}^{2}} \\right)2x}{2x\\left( x+3 \\right)\\left( x-3 \\right)}$$-\\dfrac{3.2x\\left( x+3 \\right)\\left( x-3 \\right)}{2x\\left( x+3 \\right)\\left( x-3 \\right)} $<br\/>$ =\\dfrac{3\\left( x-3 \\right)-\\left( 4-3{{x}^{2}} \\right)2x-6x\\left( {{x}^{2}}-9 \\right)}{2x\\left( x+3 \\right)\\left( x-3 \\right)} $<br\/>$ =\\dfrac{3x-9-8x+6{{x}^{3}}-6{{x}^{3}}+54x}{2x\\left( x+3 \\right)\\left( x-3 \\right)} $<br\/>$ =\\dfrac{49x-9}{2x\\left( x+3 \\right)\\left( x-3 \\right)} $ (1) <br\/> Thay $x = 1$ v\u00e0o bi\u1ec3u th\u1ee9c (1), ta \u0111\u01b0\u1ee3c: <br\/> $\\dfrac{49x-9}{2x\\left( x+3 \\right)\\left( x-3 \\right)}$$=\\dfrac{49-9}{2.4.\\left( -2 \\right)}=\\dfrac{-5}{2}$<\/span> "}]}],"id_ques":91},{"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/bai13/lv3/img\/9.jpg' \/><\/center> Cho bi\u1ec3u th\u1ee9c $\\dfrac{1}{a+3}+P=\\dfrac{6a}{3{{a}^{2}}-27}.$ Khi \u0111\u00f3 ph\u00e2n th\u1ee9c $P$ l\u00e0:","select":["A. $\\dfrac{1}{a-3}$ ","B. $\\dfrac{2}{a-3}$ ","C. $\\dfrac{a}{a+3}$","D. $\\dfrac{a}{a-3}$"],"explain":"<span class='basic_left'>Ta c\u00f3: <br\/> $\\begin{align} & \\dfrac{1}{a+3}+P=\\dfrac{6a}{3{{a}^{2}}-27} \\\\ &\\Leftrightarrow P=\\dfrac{6a}{3{{a}^{2}}-27}-\\dfrac{1}{a+3} \\\\ & =\\dfrac{6a}{3\\left( {{a}^{2}}-9 \\right)}-\\dfrac{1}{a+3} \\\\ & =\\dfrac{2a}{\\left( a+3 \\right)\\left( a-3 \\right)}-\\dfrac{1}{a+3} \\\\ & =\\dfrac{2a}{\\left( a+3 \\right)\\left( a-3 \\right)}-\\dfrac{a-3}{\\left( a+3 \\right)\\left( a-3 \\right)} \\\\ & =\\dfrac{2a-a+3}{\\left( a+3 \\right)\\left( a-3 \\right)} \\\\ & =\\dfrac{a+3}{\\left( a+3 \\right)\\left( a-3 \\right)} \\\\ & =\\dfrac{1}{a-3} \\\\ \\end{align}$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span>","column":2}]}],"id_ques":92},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["1"],["3"],["-1"],["3"],["-2"],["3"]]],"list":[{"point":10,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai13/lv3/img\/8.jpg' \/><\/center> T\u00ecm $a, b, c$ sao cho $\\dfrac{1}{{{x}^{3}}-1}$$=\\dfrac{a}{x-1}+\\dfrac{bx+c}{{{x}^{2}}+x+1}$ v\u1edbi $x\\ne \\{1\\}$ <br\/> \u0110\u00e1p \u00e1n: $a =$ <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>; $b =$ <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>; $c =$ <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","explain":"<span class='basic_left'> Ta c\u00f3: <br\/> $\\dfrac{a}{x-1}+\\dfrac{bx+c}{{{x}^{2}}+x+1}$<br\/>$=\\dfrac{a\\left( {{x}^{2}}+x+1 \\right)}{\\left( x-1 \\right)\\left( {{x}^{2}}+x+1 \\right)}$$+\\dfrac{\\left( bx+c \\right)\\left( x-1 \\right)}{\\left( x-1 \\right)\\left( {{x}^{2}}+x+1 \\right)} $<br\/>$ =\\dfrac{a\\left( {{x}^{2}}+x+1 \\right)+\\left( bx+c \\right)\\left( x-1 \\right)}{\\left( x-1 \\right)\\left( {{x}^{2}}+x+1 \\right)} $<br\/>$ =\\dfrac{a\\,{{x}^{2}}+a\\,x+a+b{{x}^{2}}-bx+cx-c}{\\left( x-1 \\right)\\left( {{x}^{2}}+x+1 \\right)} $<br\/>$ =\\dfrac{\\left( a+b \\right){{x}^{2}}+\\left( a-b+c \\right)x+a-c}{{{x}^{3}}-1} $<br\/> \u0110\u1ed3ng nh\u1ea5t h\u1ec7 s\u1ed1 hai t\u1eed th\u1ee9c: $1\\equiv \\left( a+b \\right){{x}^{2}}$$+\\left( a-b+c \\right)x+a-c$ , ta \u0111\u01b0\u1ee3c:<br\/> $\\left\\{ \\begin{aligned} & a+b=0 \\\\ & a-b+c=0 \\\\ & a-c=1 \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & a=\\dfrac{1}{3} \\\\ & b=\\dfrac{-1}{3} \\\\ & c=\\dfrac{-2}{3} \\\\ \\end{aligned} \\right.$<\/span> "}]}],"id_ques":93},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["3"],["5"],["7"],["5"]]],"list":[{"point":10,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai13/lv3/img\/4.jpg' \/><\/center> T\u00ecm $m, n$ sao cho $\\dfrac{2x-1}{{{x}^{2}}+x-6}$$=\\dfrac{m}{x-2}+\\dfrac{n}{x+3}$ v\u1edbi $x\\ne \\{-3;2\\}$ <br\/> \u0110\u00e1p \u00e1n: $m =$ <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>; $n =$ <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","explain":"<span class='basic_left'> Ta c\u00f3: <br\/> $\\begin{align} &\\dfrac{m}{x-2}+\\dfrac{n}{x+3} \\\\&=\\dfrac{m\\left( x+3 \\right)}{\\left( x-2 \\right)\\left( x+3 \\right)}+\\dfrac{n\\left( x-2 \\right)}{\\left( x-2 \\right)\\left( x+3 \\right)} \\\\ & =\\dfrac{m\\left( x+3 \\right)+n\\left( x-2 \\right)}{\\left( x-2 \\right)\\left( x+3 \\right)} \\\\ & =\\dfrac{mx+3m+nx-2n}{\\left( x-2 \\right)\\left( x+3 \\right)} \\\\ & =\\dfrac{\\left( m+n \\right)x+3m-2n}{{{x}^{2}}+x-6} \\\\ \\end{align}$<br\/> \u0110\u1ed3ng nh\u1ea5t hai t\u1eed th\u1ee9c: $2x-1$$\\equiv \\left( m+n \\right)x+3m-2n$ , ta \u0111\u01b0\u1ee3c: <br\/> $\\left\\{ \\begin{aligned} & m+n=2 \\\\ & 3m-2n=-1 \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & m=\\dfrac{3}{5} \\\\ & n=\\dfrac{7}{5} \\\\ \\end{aligned} \\right.$<\/span> "}]}],"id_ques":94},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t.<br\/> K\u1ebft qu\u1ea3 ph\u00e9p t\u00ednh d\u01b0\u1edbi \u0111\u00e2y \u0111\u00fang hay sai","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai13/lv3/img\/2.jpg' \/><\/center> $\\dfrac{{{x}^{4}}}{1-x}+{{x}^{3}}+{{x}^{2}}+x+1=\\dfrac{x+1}{1-x}$ ","select":["\u0110\u00fang","Sai"],"explain":"<span class='basic_left'> Ta c\u00f3: <br\/> V\u1ebf tr\u00e1i $= \\dfrac{{{x}^{4}}}{1-x}+{{x}^{3}}+{{x}^{2}}+x+1 $<br\/>$ =\\dfrac{{{x}^{4}}}{1-x}$$+\\dfrac{\\left( {{x}^{3}}+{{x}^{2}}+x+1 \\right)\\left( 1-x \\right)}{1-x} $<br\/>$ =\\dfrac{{{x}^{4}}+\\left( {{x}^{3}}+{{x}^{2}}+x+1 \\right)\\left( 1-x \\right)}{1-x} $<br\/>$ =\\dfrac{{{x}^{4}}+{{x}^{3}}+{{x}^{2}}+x+1-{{x}^{4}}-{{x}^{3}}-{{x}^{2}}-x}{1-x} $<br\/>$ =\\dfrac{1}{1-x} \\ne $ V\u1ebf ph\u1ea3i <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 Sai.<\/span>","column":2}]}],"id_ques":95},{"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/bai13/lv3/img\/10.jpg' \/><\/center> K\u1ebft qu\u1ea3 c\u1ee7a ph\u00e9p to\u00e1n $\\dfrac{x+1}{x-3}+\\dfrac{1-x}{x+3}$$-\\dfrac{2x\\left( 1-x \\right)}{9-{{x}^{2}}} $ l\u00e0: ","select":["A. $\\dfrac{2}{x-3}$ ","B. $\\dfrac{2x\\left( x-5 \\right)}{\\left( x+3 \\right)\\left( x-3 \\right)}$","C. $\\dfrac{x-5}{\\left( x+3 \\right)\\left( x-3 \\right)}$","D. $\\dfrac{-2x\\left( x-5 \\right)}{\\left( x+3 \\right)\\left( x-3 \\right)}$"],"explain":"<span class='basic_left'> Ta c\u00f3:<br\/> $ \\dfrac{x+1}{x-3}+\\dfrac{1-x}{x+3}-\\dfrac{2x\\left( 1-x \\right)}{9-{{x}^{2}}} $<br\/>$=\\dfrac{\\left( x+1 \\right)\\left( x+3 \\right)}{\\left( x+3 \\right)\\left( x-3 \\right)}$$+\\dfrac{\\left( 1-x \\right)\\left( x-3 \\right)}{\\left( x+3 \\right)\\left( x-3 \\right)}$$+\\dfrac{2x\\left( 1-x \\right)}{\\left( x+3 \\right)\\left( x-3 \\right)} $<br\/>$ =\\dfrac{\\left( x+1 \\right)\\left( x+3 \\right)+\\left( 1-x \\right)\\left( x-3 \\right)+2x\\left( 1-x \\right)}{\\left( x+3 \\right)\\left( x-3 \\right)} $<br\/>$ =\\dfrac{{{x}^{2}}+4x+3+x-3-{{x}^{2}}+3x+2x-2{{x}^{2}}}{\\left( x+3 \\right)\\left( x-3 \\right)} $<br\/>$ =\\dfrac{-2{{x}^{2}}+10x}{\\left( x+3 \\right)\\left( x-3 \\right)} $<br\/>$ =\\dfrac{-2x\\left( x-5 \\right)}{\\left( x+3 \\right)\\left( x-3 \\right)} $ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D. <\/span><\/span> ","column":2}]}],"id_ques":96},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng trong ph\u00e9p c\u1ed9ng (tr\u1eeb) c\u00e1c ph\u00e2n th\u1ee9c sau","title_trans":"","temp":"fill_the_blank","correct":[[["3"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai13/lv3/img\/11.jpg' \/><\/center> ${{x}^{2}}+1-\\dfrac{{{x}^{4}}-3{{x}^{2}}+2}{{{x}^{2}}-1}$$=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$ ","explain":"<span class='basic_left'> Ta c\u00f3:<br\/> $\\begin{align} & {{x}^{2}}+1-\\dfrac{{{x}^{4}}-3{{x}^{2}}+2}{{{x}^{2}}-1}\\\\ &=\\dfrac{\\left( {{x}^{2}}+1 \\right)\\left( {{x}^{2}}-1 \\right)}{{{x}^{2}}-1}-\\dfrac{{{x}^{4}}-3{{x}^{2}}+2}{{{x}^{2}}-1} \\\\ & =\\dfrac{{{x}^{4}}-1}{{{x}^{2}}-1}-\\dfrac{{{x}^{4}}-3{{x}^{2}}+2}{{{x}^{2}}-1} \\\\ & =\\dfrac{{{x}^{4}}-1-{{x}^{4}}+3{{x}^{2}}-2}{{{x}^{2}}-1} \\\\ & =\\dfrac{3{{x}^{2}}-3}{{{x}^{2}}-1} \\\\ & =\\dfrac{3\\left( {{x}^{2}}-1 \\right)}{{{x}^{2}}-1} \\\\ & =3 \\\\ \\end{align}$ <br\/> <span class='basic_pink'> K\u1ebft qu\u1ea3 c\u1ee7a ph\u00e9p to\u00e1n tr\u00ean l\u00e0 $3$ <\/span><\/span> "}]}],"id_ques":97},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng trong ph\u00e9p c\u1ed9ng (tr\u1eeb) c\u00e1c ph\u00e2n th\u1ee9c sau","title_trans":"","temp":"fill_the_blank","correct":[[["2xy"],["x+y","y+x"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai13/lv3/img\/9.jpg' \/><\/center> $x+y-\\dfrac{{{x}^{2}}+{{y}^{2}}}{x+y}=\\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$ ","explain":"<span class='basic_left'> Ta c\u00f3:<br\/> $\\begin{align} &x+y-\\dfrac{{{x}^{2}}+{{y}^{2}}}{x+y} \\\\& =\\dfrac{\\left( x+y \\right)\\left( x+y \\right)}{x+y}-\\dfrac{{{x}^{2}}+{{y}^{2}}}{x+y} \\\\ & =\\dfrac{{{x}^{2}}+2xy+{{y}^{2}}}{x+y}-\\dfrac{{{x}^{2}}+{{y}^{2}}}{x+y} \\\\ & =\\dfrac{{{x}^{2}}+2xy+{{y}^{2}}-{{x}^{2}}-{{y}^{2}}}{x+y} \\\\ & =\\dfrac{2xy}{x+y} \\\\ \\end{align}$ <br\/> <span class='basic_pink'> K\u1ebft qu\u1ea3 c\u1ee7a ph\u00e9p t\u00ednh tr\u00ean l\u00e0 $\\dfrac{2xy}{x+y}$ <\/span><\/span> "}]}],"id_ques":98},{"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/bai13/lv3/img\/8.jpg' \/><\/center> Quy \u0111\u1ed3ng m\u1eabu th\u1ee9c c\u00e1c ph\u00e2n th\u1ee9c $\\dfrac{2x}{x^2-8x+16}$; $\\dfrac{x}{3x^2-12x}$ v\u00e0 $\\dfrac{x+1}{x^2-16}$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0:","select":["A. $\\dfrac{2x\\left( x+4 \\right)}{{{x}^{2}}-8x+16};\\dfrac{x\\left( x-4 \\right)}{3{{x}^{2}}-12x};\\dfrac{\\left( x+1 \\right)\\left( x-4 \\right)}{{{x}^{2}}-16}$ ","B. $\\dfrac{6{{x}^{2}}\\left( x+4 \\right)}{3x\\left( x+4 \\right){{\\left( x-4 \\right)}^{2}}};\\,\\,\\dfrac{x\\left( {{x}^{2}}-16 \\right)}{3x\\left( x+4 \\right){{\\left( x-4 \\right)}^{2}}};\\,\\,\\dfrac{3x\\left( x-4 \\right)\\left( x+1 \\right)}{3x\\left( x+4 \\right){{\\left( x-4 \\right)}^{2}}}$ ","C. $\\dfrac{2x\\left( x+4 \\right)}{3x\\left( x+4 \\right){{\\left( x-4 \\right)}^{2}}};\\,\\,\\,\\dfrac{x\\left( {{x}^{2}}-4 \\right)}{3x\\left( x+4 \\right){{\\left( x-4 \\right)}^{2}}};\\,\\,\\dfrac{\\left( x+1 \\right)\\left( x-4 \\right)}{3x\\left( x+4 \\right){{\\left( x-4 \\right)}^{2}}}$ "],"explain":"<span class='basic_left'>Ta ph\u00e2n t\u00edch c\u00e1c m\u1eabu th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed:<br\/> $\\begin{align} & {{x}^{2}}-8x+16={{\\left( x-4 \\right)}^{2}} \\\\ & 3{{x}^{2}}-12x=3x\\left( x-4 \\right) \\\\ & {{x}^{2}}-16=\\left( x+4 \\right)\\left( x-4 \\right) \\\\ \\end{align}$<br\/> M\u1eabu th\u1ee9c chung: $3x\\left( x+4 \\right){{\\left( x-4 \\right)}^{2}}$ <br\/> Nh\u00e2n t\u1eed ph\u1ee5 c\u1ee7a m\u1eabu th\u1ee9c ${{\\left( x-4 \\right)}^{2}}$ l\u00e0: $3x\\left( x+4 \\right)$ . Do \u0111\u00f3:<br\/> $\\dfrac{2x}{{{x}^{2}}-8x+16}=\\dfrac{2x}{{{\\left( x-4 \\right)}^{2}}}$$=\\dfrac{2x.3x\\left( x+4 \\right)}{3x\\left( x+4 \\right){{\\left( x-4\\right)}^{2}}}$$=\\dfrac{6{{x}^{2}}\\left( x+4 \\right)}{3x\\left( x+4 \\right){{\\left( x-4 \\right)}^{2}}}$ <br\/> Nh\u00e2n t\u1eed ph\u1ee5 c\u1ee7a m\u1eabu th\u1ee9c $3x\\left( x-4 \\right)$ l\u00e0: $\\left( x+4 \\right)\\left( x-4 \\right)={{x}^{2}}-16$ . Do \u0111\u00f3:<br\/> $\\dfrac{x}{3{{x}^{2}}-12x}=\\dfrac{x}{3x\\left( x-4 \\right)}$$=\\dfrac{x\\left( {{x}^{2}}-16 \\right)}{3x\\left( x+4 \\right){{\\left( x-4 \\right)}^{2}}}$ <br\/> Nh\u00e2n t\u1eed ph\u1ee5 c\u1ee7a m\u1eabu th\u1ee9c $\\left( x+4 \\right)\\left( x-4 \\right)$ l\u00e0: $3x\\left( x-4 \\right)$ . Do \u0111\u00f3:<br\/> $\\dfrac{x+1}{{{x}^{2}}-16}=\\dfrac{x+1}{\\left( x+4 \\right)\\left( x-4 \\right)}$$=\\dfrac{3x\\left( x+1 \\right)\\left( x-4 \\right)}{3x\\left( x+4 \\right){{\\left( x-4 \\right)}^{2}}}$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span>","column":1}]}],"id_ques":99},{"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/bai13/lv3/img\/2.jpg' \/><\/center> M\u1eabu th\u1ee9c chung c\u1ee7a hai ph\u00e2n th\u1ee9c $\\dfrac{1}{x^2-2xy+y^2-z^2}$ v\u00e0 $\\dfrac{1}{x-y-z}$ l\u00e0","select":["A. $x-y+z$ ","B. $x+y+z$ ","C. $x-y-z$","D. $(x-y+z)(x-y-z)$"],"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 c\u00e1c m\u1eabu th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed <br\/> <b> B\u01b0\u1edbc 2:<\/b> L\u1ea5y t\u00edch c\u1ee7a BCNN c\u00e1c h\u1ec7 s\u1ed1 c\u1ee7a c\u00e1c th\u1eeba s\u1ed1 chung v\u00e0 ri\u00eang c\u00f3 m\u1eb7t trong m\u1eabu th\u1ee9c v\u1edbi s\u1ed1 m\u0169 cao nh\u1ea5t <br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'> Ph\u00e2n t\u00edch m\u1eabu th\u1ee9c c\u1ee7a c\u00e1c ph\u00e2n th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed:<br\/> $x^2-2xy+y^2-z^2=(x-y)^2-z^2$$=(x-y+z)(x-y-z)$ <br\/> M\u1eabu th\u1ee9c th\u1ee9 nh\u1ea5t l\u00e0 $(x-y+z)(x-y-z)$ <br\/> M\u1eabu th\u1ee9c th\u1ee9 hai l\u00e0 $x-y-z$ <br\/> V\u1eady m\u1eabu th\u1ee9c chung c\u1ee7a $\\dfrac{1}{x^2-2xy+y^2-z^2}$ v\u00e0 $\\dfrac{1}{x-y-z}$ l\u00e0: $(x-y+z)(x-y-z)$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span>","column":2}]}],"id_ques":100}],"lesson":{"save":0,"level":3}}