{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-1"],["2"]]],"list":[{"point":5,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai13/lv2/img\/12.jpg' \/><\/center> T\u00ecm $m, n$ sao cho $\\dfrac{x+1}{(x+2)(x+3)}=\\dfrac{m}{x+2}+\\dfrac{n}{x+3}$ v\u1edbi $x\\ne \\{-3;-2\\}$ <br\/> \u0110\u00e1p \u00e1n: $m =$ _input_ ; $n =$ _input_","hint":" Quy \u0111\u1ed3ng m\u1eabu th\u1ee9c v\u00e0 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $\\dfrac{m}{x+2}+\\dfrac{n}{x+3}$<br\/> S\u1eed d\u1ee5ng \u0111\u1ed3ng nh\u1ea5t h\u1ec7 s\u1ed1 hai t\u1eed th\u1ee9c \u0111\u1ec3 t\u00ecm $m, n.$ ","explain":"<span class='basic_left'> Ta c\u00f3: <br\/> $\\begin{align} &\\dfrac{m}{x+2}+\\dfrac{n}{x+3} \\\\& =\\dfrac{m\\left( x+3 \\right)}{\\left( x+2 \\right)\\left( x+3 \\right)}+\\dfrac{n\\left( x+2 \\right)}{\\left( x+2 \\right)\\left( x+3 \\right)} \\\\ & =\\dfrac{mx+3m+nx+2n}{\\left( x+2 \\right)\\left( x+3 \\right)} \\\\ & =\\dfrac{\\left( m+n \\right)x+3m+2n}{\\left( x+2 \\right)\\left( x+3 \\right)} \\\\ \\end{align}$<br\/> \u0110\u1ed3ng nh\u1ea5t h\u1ec7 s\u1ed1 hai t\u1eed th\u1ee9c: $x+1\\equiv \\left( m+n \\right)x+3m+2n,$ ta \u0111\u01b0\u1ee3c:<br\/> $\\left\\{ \\begin{aligned} & m+n=1 \\\\ & 3m+2n=1 \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & m=-1 \\\\ & n=2 \\\\ \\end{aligned} \\right.$<\/span> "}]}],"id_ques":71},{"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":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai13/lv2/img\/11.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $ \\dfrac{x+1}{2x+6}+\\dfrac{2x+3}{x\\left( x+3 \\right)}$ t\u1ea1i $x=-2$ l\u00e0 _input_ ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x \\ne \\{-3;0\\}$ <br\/> Ta c\u00f3: <br\/> $\\begin{align} &\\dfrac{x+1}{2x+6}+\\dfrac{2x+3}{x\\left( x+3 \\right)} \\\\&=\\dfrac{x+1}{2\\left( x+3 \\right)}+\\dfrac{2x+3}{x\\left( x+3 \\right)} \\\\ & =\\dfrac{\\left( x+1 \\right)x}{2x\\left( x+3 \\right)}+\\dfrac{\\left( 2x+3 \\right)2}{2x\\left( x+3 \\right)} \\\\ & =\\dfrac{{{x}^{2}}+x}{2x\\left( x+3 \\right)}+\\dfrac{4x+6}{2x\\left( x+3 \\right)} \\\\ & =\\dfrac{{{x}^{2}}+x+4x+6}{2x\\left( x+3 \\right)} \\\\ & =\\dfrac{{{x}^{2}}+5x+6}{2x\\left( x+3 \\right)} \\\\ & =\\dfrac{{{x}^{2}}+3x+2x+6}{2x\\left( x+3 \\right)} \\\\ & =\\dfrac{x\\left( x+3 \\right)+2\\left( x+3 \\right)}{2x\\left( x+3 \\right)} \\\\ & =\\dfrac{\\left( x+3 \\right)\\left( x+2 \\right)}{2x\\left( x+3 \\right)} \\\\ & =\\dfrac{x+2}{2x} \\,\\, (1) \\\\ \\end{align}$ <br\/> Thay $x=-2$ v\u00e0o (1), ta \u0111\u01b0\u1ee3c:<br\/> $\\dfrac{x+2}{2x}=\\dfrac{-2+2}{2.(-2)}=0$<\/span> "}]}],"id_ques":72},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-12"]]],"list":[{"point":5,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai13/lv2/img\/10.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $\\dfrac{4{{x}^{2}}-3x+17}{{{x}^{3}}-1}+\\dfrac{2x-1}{{{x}^{2}}+x+1}$$+\\dfrac{6}{1-x}$ t\u1ea1i $x=-1$ l\u00e0 _input_ ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x \\ne 1$ <br\/> Ta c\u00f3: <br\/> $ \\dfrac{4{{x}^{2}}-3x+17}{{{x}^{3}}-1}+\\dfrac{2x-1}{{{x}^{2}}+x+1}$$+\\dfrac{6}{1-x} $ <br\/> $ =\\dfrac{4{{x}^{2}}-3x+17}{{{x}^{3}}-1}+\\dfrac{2x-1}{{{x}^{2}}+x+1}$$-\\dfrac{6}{x-1} $ <br\/> $ =\\dfrac{4{{x}^{2}}-3x+17}{\\left( x-1 \\right)\\left( {{x}^{2}}+x+1 \\right)}$$+\\dfrac{\\left( 2x-1 \\right)\\left( x-1 \\right)}{\\left( x-1 \\right)\\left( {{x}^{2}}+x+1 \\right)}$$-\\dfrac{6\\left( {{x}^{2}}+x+1 \\right)}{\\left( x-1 \\right)\\left( {{x}^{2}}+x+1 \\right)} $ <br\/> $ =\\dfrac{4{{x}^{2}}-3x+17+\\left( 2x-1 \\right)\\left( x-1 \\right)-6\\left( {{x}^{2}}+x+1 \\right)}{\\left( x-1 \\right)\\left( {{x}^{2}}+x+1 \\right)} $ <br\/> $ =\\dfrac{4{{x}^{2}}-3x+17+2{{x}^{2}}-3x+1-6{{x}^{2}}-6x-6}{\\left( x-1 \\right)\\left( {{x}^{2}}+x+1 \\right)} $ <br\/> $ =\\dfrac{-12x+12}{\\left( x-1 \\right)\\left( {{x}^{2}}+x+1 \\right)} $ <br\/> $ =\\dfrac{-12\\left( x-1 \\right)}{\\left( x-1 \\right)\\left( {{x}^{2}}+x+1 \\right)} $ <br\/> $ =\\dfrac{-12}{{{x}^{2}}+x+1}$ <br\/> Thay $x=-1$ v\u00e0o bi\u1ec3u th\u1ee9c r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c: <br\/> $\\dfrac{-12}{x^2+x+1}=\\dfrac{-12}{1-1+1}=-12$ <\/span> "}]}],"id_ques":73},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["5"],["18"]]],"list":[{"point":5,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai13/lv2/img\/9.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $\\dfrac{1}{x+2}+\\dfrac{3}{{{x}^{2}}-4}$$+\\dfrac{x-14}{\\left( {{x}^{2}}+4x+4 \\right)\\left( x-2 \\right)}$ t\u1ea1i $x=4$ l\u00e0 <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x \\ne \\{-2;2\\}$ <br\/> Ta c\u00f3: <br\/> $ \\dfrac{1}{x+2}+\\dfrac{3}{{{x}^{2}}-4}$$+\\dfrac{x-14}{\\left( {{x}^{2}}+4x+4 \\right)\\left( x-2 \\right)}$<br\/>$=\\dfrac{1}{x+2}+\\dfrac{3}{\\left( x+2 \\right)\\left( x-2 \\right)}$$+\\dfrac{x-14}{{{\\left( x+2 \\right)}^{2}}\\left( x-2 \\right)} $<br\/>$ =\\dfrac{\\left( x+2 \\right)\\left( x-2 \\right)}{{{\\left( x+2 \\right)}^{2}}\\left( x-2 \\right)}$$+\\dfrac{3\\left( x+2 \\right)}{{{\\left( x+2 \\right)}^{2}}\\left( x-2 \\right)}$$+\\dfrac{x-14}{{{\\left( x+2 \\right)}^{2}}\\left( x-2 \\right)} $<br\/>$=\\dfrac{{{x}^{2}}-4}{{{\\left( x+2 \\right)}^{2}}\\left( x-2 \\right)}$$+\\dfrac{3x+6}{{{\\left( x+2 \\right)}^{2}}\\left( x-2 \\right)}$$+\\dfrac{x-14}{{{\\left( x+2 \\right)}^{2}}\\left( x-2 \\right)} $<br\/>$ =\\dfrac{{{x}^{2}}-4+3x+6+x-14}{{{\\left( x+2 \\right)}^{2}}\\left( x-2 \\right)} $<br\/>$ =\\dfrac{{{x}^{2}}+4x-12}{{{\\left( x+2 \\right)}^{2}}\\left( x-2 \\right)} $<br\/>$ =\\dfrac{{{x}^{2}}+6x-2x-12}{{{\\left( x+2 \\right)}^{2}}\\left( x-2 \\right)} $<br\/>$=\\dfrac{x\\left( x+6 \\right)-2\\left( x+6 \\right)}{{{\\left( x+2 \\right)}^{2}}\\left( x-2 \\right)} $<br\/>$=\\dfrac{\\left( x+6 \\right)\\left( x-2 \\right)}{{{\\left( x+2 \\right)}^{2}}\\left( x-2 \\right)} $<br\/>$ =\\dfrac{x+6}{{{\\left( x+2 \\right)}^{2}}} $ <br\/> Thay $x = 4$ v\u00e0o bi\u1ec3u th\u1ee9c r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c:<br\/> $\\dfrac{x+6}{{{\\left( x+2 \\right)}^{2}}}=\\dfrac{4+6}{{{\\left( 4+2 \\right)}^{2}}}=\\dfrac{10}{36}=\\dfrac{5}{18}$<\/span> "}]}],"id_ques":74},{"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":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai13/lv2/img\/8.jpg' \/><\/center> T\u00ecm $x,$ bi\u1ebft $\\dfrac{2x+1}{{{\\left( x-1 \\right)}^{2}}}-\\dfrac{2x+3}{{{x}^{2}}-1}=0$<br\/> \u0110\u00e1p \u00e1n: $x =$ _input_ ","hint":"Th\u1ef1c hi\u1ec7n quy \u0111\u1ed3ng v\u00e0 r\u00fat g\u1ecdn v\u1ebf tr\u00e1i. ","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 c\u1ee7a $x$ \u0111\u1ec3 c\u00e1c ph\u00e2n th\u1ee9c c\u00f3 ngh\u0129a.<br\/> <b> B\u01b0\u1edbc 2:<\/b> Th\u1ef1c hi\u1ec7n quy \u0111\u1ed3ng v\u00e0 r\u00fat g\u1ecdn v\u1ebf tr\u00e1i \u0111\u1ec3 t\u00ecm $x.$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> \u0110i\u1ec1u ki\u1ec7n: $x\\ne \\{-1;1\\}$.<br\/> Ta c\u00f3: <br\/> $\\begin{align} & \\dfrac{2x+1}{{{\\left( x-1 \\right)}^{2}}}-\\dfrac{2x+3}{{{x}^{2}}-1}=0 \\\\ &\\Leftrightarrow \\dfrac{2x+1}{{{\\left( x-1 \\right)}^{2}}}-\\dfrac{2x+3}{\\left( x+1 \\right)\\left( x-1 \\right)}=0 \\\\ & \\Leftrightarrow \\dfrac{\\left( 2x+1 \\right)\\left( x+1 \\right)}{\\left( x+1 \\right){{\\left( x-1 \\right)}^{2}}}-\\dfrac{\\left( 2x+3 \\right)\\left( x-1 \\right)}{\\left( x+1 \\right){{\\left( x-1 \\right)}^{2}}}=0 \\\\ &\\Leftrightarrow \\dfrac{2{{x}^{2}}+3x+1}{\\left( x+1 \\right){{\\left( x-1 \\right)}^{2}}}-\\dfrac{2{{x}^{2}}+x-3}{\\left( x+1 \\right){{\\left( x-1 \\right)}^{2}}}=0 \\\\ &\\Leftrightarrow \\dfrac{2{{x}^{2}}+3x+1-2{{x}^{2}}-x+3}{\\left( x+1 \\right){{\\left( x-1 \\right)}^{2}}}=0 \\\\ &\\Leftrightarrow \\dfrac{2x+4}{\\left( x+1 \\right){{\\left( x-1 \\right)}^{2}}}=0 \\\\ & \\Rightarrow 2x+4=0 \\\\ & \\Leftrightarrow x=-2\\,\\,\\,\\,\\left( \\text{th\u1ecfa m\u00e3n} \\right) \\\\ \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-2$. <\/span><\/span> "}]}],"id_ques":75},{"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":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai13/lv2/img\/5.jpg' \/><\/center> T\u00ecm $x,$ bi\u1ebft $\\dfrac{2{{x}^{2}}-3x+2}{{{x}^{3}}-1}+\\dfrac{4x-{{x}^{2}}-1}{{{x}^{3}}-1}=1$<br\/> \u0110\u00e1p \u00e1n: $x =$ _input_ ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x\\ne 1$.<br\/> Ta c\u00f3: <br\/> $\\begin{align} & \\dfrac{2{{x}^{2}}-3x+2}{{{x}^{3}}-1}+\\dfrac{4x-{{x}^{2}}-1}{{{x}^{3}}-1}=1 \\\\ &\\Leftrightarrow \\dfrac{2{{x}^{2}}-3x+2+4x-{{x}^{2}}-1}{{{x}^{3}}-1}=1 \\\\ & \\Leftrightarrow\\dfrac{{{x}^{2}}+x+1}{{{x}^{3}}-1}=1 \\\\ &\\Leftrightarrow \\dfrac{{{x}^{2}}+x+1}{\\left( x-1 \\right)\\left( {{x}^{2}}+x+1 \\right)}=1 \\\\ &\\Leftrightarrow \\dfrac{1}{x-1}=1 \\\\ &\\Leftrightarrow x-1=1 \\\\ &\\Leftrightarrow x=2\\,\\,\\,\\left( th\u1ecfa \\, m\u00e3n \\right) \\\\ \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $2$. <\/span><\/span> "}]}],"id_ques":76},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["9"],["2"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai13/lv2/img\/4.jpg' \/><\/center> T\u00ecm $x$ bi\u1ebft $\\dfrac{1}{\\left( x-5 \\right)\\left( x-4 \\right)}+\\dfrac{1}{\\left( x-5 \\right)\\left( x-6 \\right)}$$=\\dfrac{-2}{\\left( x-5 \\right)\\left( x-6 \\right)}$<br\/> \u0110\u00e1p \u00e1n: $x=$ <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x\\ne \\{4;5;6\\}$.<br\/> Ta c\u00f3: <br\/> $ \\dfrac{1}{\\left( x-5 \\right)\\left( x-4 \\right)}+\\dfrac{1}{\\left( x-5 \\right)\\left( x-6 \\right)}$$=\\dfrac{-2}{\\left( x-5 \\right)\\left( x-6 \\right)} $ <br\/> $\\Leftrightarrow \\dfrac{1}{\\left( x-5 \\right)\\left( x-4 \\right)}+\\dfrac{1}{\\left( x-5 \\right)\\left( x-6 \\right)}$$-\\dfrac{-2}{\\left( x-5 \\right)\\left( x-6 \\right)}=0 $ <br\/> $\\Leftrightarrow \\dfrac{x-6}{\\left( x-4 \\right)\\left( x-5 \\right)\\left( x-6 \\right)}$$+\\dfrac{x-4}{\\left( x-4 \\right)\\left( x-5 \\right)\\left( x-6 \\right)}$$-\\dfrac{-2\\left( x-4 \\right)}{\\left( x-4 \\right)\\left( x-5 \\right)\\left( x-6 \\right)}=0 $ <br\/> $\\Leftrightarrow \\dfrac{x-6+x-4+2x-8}{\\left( x-4 \\right)\\left( x-5 \\right)\\left( x-6 \\right)}=0 $ <br\/> $\\Leftrightarrow \\dfrac{4x-18}{\\left( x-4 \\right)\\left( x-5 \\right)\\left( x-6 \\right)}=0 $ <br\/> $\\Rightarrow 4x-18=0 $ <br\/> $\\Leftrightarrow x=\\dfrac{18}{4}=\\dfrac{9}{2} \\,(\\text{th\u1ecfa m\u00e3n})$<\/span> "}]}],"id_ques":77},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t.<br\/> K\u1ebft qu\u1ea3 ph\u00e9p t\u00ednh d\u01b0\u1edbi \u0111\u00e2y \u0111\u00fang hay sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai13/lv2/img\/3.jpg' \/><\/center> $ x-\\dfrac{xy}{x+y}-\\dfrac{{{x}^{3}}}{{{x}^{2}}-{{y}^{2}}}$$=\\dfrac{-{{x}^{2}}y}{{{x}^{2}}-{{y}^{2}}}$ ","select":["\u0110\u00fang","Sai"],"hint":" Ta th\u1ef1c hi\u1ec7n r\u00fat g\u1ecdn v\u1ebf tr\u00e1i r\u1ed3i so s\u00e1nh k\u1ebft qu\u1ea3 v\u1edbi v\u1ebf ph\u1ea3i. ","explain":"<span class='basic_left'> Ta c\u00f3: <br\/> V\u1ebf tr\u00e1i = $x-\\dfrac{xy}{x+y}-\\dfrac{{{x}^{3}}}{{{x}^{2}}-{{y}^{2}}}$<br\/>$=x-\\dfrac{xy}{x+y}-\\dfrac{{{x}^{3}}}{\\left( x+y \\right)\\left( x-y \\right)} $<br\/>$ =\\dfrac{x\\left( {{x}^{2}}-{{y}^{2}} \\right)}{\\left( x+y \\right)\\left( x-y \\right)}$$-\\dfrac{xy\\left( x-y \\right)}{\\left( x+y \\right)\\left( x-y \\right)}$$-\\dfrac{{{x}^{3}}}{\\left( x+y \\right)\\left( x-y \\right)} $<br\/>$ =\\dfrac{{{x}^{3}}-x{{y}^{2}}}{\\left( x+y \\right)\\left( x-y \\right)}$$-\\dfrac{{{x}^{2}}y-x{{y}^{2}}}{\\left( x+y \\right)\\left( x-y \\right)}$$-\\dfrac{{{x}^{3}}}{\\left( x+y \\right)\\left( x-y \\right)} $<br\/>$ =\\dfrac{{{x}^{3}}-x{{y}^{2}}-{{x}^{2}}y+x{{y}^{2}}-{{x}^{3}}}{\\left( x+y \\right)\\left( x-y \\right)} $<br\/>$ =\\dfrac{-{{x}^{2}}y}{{{x}^{2}}-{{y}^{2}}} $ = V\u1ebf ph\u1ea3i <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang.<\/span>","column":2}]}],"id_ques":78},{"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":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai13/lv2/img\/2.jpg' \/><\/center> $\\dfrac{1}{x}+\\dfrac{1}{x+1}+\\dfrac{1-2x}{x\\left( x+1 \\right)}=\\dfrac{3-2x}{x(x+1)}$ ","select":["\u0110\u00fang","Sai"],"explain":"<span class='basic_left'> Ta c\u00f3: <br\/> V\u1ebf tr\u00e1i = $\\dfrac{1}{x}+\\dfrac{1}{x+1}+\\dfrac{1-2x}{x\\left( x+1 \\right)} $<br\/>$=\\dfrac{1.\\left( x+1 \\right)}{x\\left( x+1 \\right)}+\\dfrac{1.x}{x\\left( x+1 \\right)}+\\dfrac{1-2x}{x\\left( x+1 \\right)}$<br\/>$ =\\dfrac{x+1}{x\\left( x+1 \\right)}+\\dfrac{x}{x\\left( x+1 \\right)}+\\dfrac{1-2x}{x\\left( x+1 \\right)} $<br\/>$ =\\dfrac{x+1+x+1-2x}{x\\left( x+1 \\right)} $<br\/>$ =\\dfrac{2}{x\\left( x+1 \\right)}\\ne$ V\u1ebf ph\u1ea3i <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 Sai.<\/span>","column":2}]}],"id_ques":79},{"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":[[1]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai13/lv2/img\/16.jpg' \/><\/center> $\\dfrac{{{x}^{3}}}{1-x}+{{x}^{2}}+x+1=\\dfrac{1}{1-x}$ ","select":["\u0110\u00fang","Sai"],"explain":"<span class='basic_left'> Ta c\u00f3: <br\/> V\u1ebf tr\u00e1i = $\\dfrac{{{x}^{3}}}{1-x}+{{x}^{2}}+x+1$<br\/>$=\\dfrac{{{x}^{3}}}{1-x}+\\dfrac{\\left( {{x}^{2}}+x+1 \\right)\\left( 1-x \\right)}{1-x} $<br\/>$ =\\dfrac{{{x}^{3}}+1-{{x}^{3}}}{1-x} $<br\/>$ =\\dfrac{1}{1-x}$ = V\u1ebf ph\u1ea3i <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang.<\/span>","column":2}]}],"id_ques":80},{"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/bai13/lv2/img\/13.jpg' \/><\/center> K\u1ebft qu\u1ea3 c\u1ee7a ph\u00e9p tr\u1eeb $\\dfrac{a+b}{a}-\\dfrac{a}{a-b}-\\dfrac{{{b}^{2}}}{{{a}^{2}}-ab}$ l\u00e0: ","select":["A. $\\dfrac{2a}{a(a-b)}$ ","B. $0$","C. $\\dfrac{b^2}{a(a-b)}$","D. $\\dfrac{-2b^2}{a(a-b)}$"],"explain":"<span class='basic_left'> Ta c\u00f3:<br\/> $\\begin{align} &\\dfrac{a+b}{a}-\\dfrac{a}{a-b}-\\dfrac{{{b}^{2}}}{{{a}^{2}}-ab}\\\\&=\\dfrac{a+b}{a}-\\dfrac{a}{a-b}-\\dfrac{{{b}^{2}}}{a\\left( a-b \\right)} \\\\ & =\\dfrac{\\left( a+b \\right)\\left( a-b \\right)}{a\\left( a-b \\right)}-\\dfrac{a.a}{a\\left( a-b \\right)}-\\dfrac{{{b}^{2}}}{a\\left( a-b \\right)} \\\\ & =\\dfrac{{{a}^{2}}-{{b}^{2}}-{{a}^{2}}-{{b}^{2}}}{a\\left( a-b \\right)} \\\\ & =\\dfrac{-2{{b}^{2}}}{a\\left( a-b \\right)} \\\\ \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D. <\/span><\/span> ","column":2}]}],"id_ques":81},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng trong ph\u00e9p tr\u1eeb c\u00e1c ph\u00e2n th\u1ee9c sau","title_trans":"","temp":"fill_the_blank","correct":[[["-6m"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai13/lv2/img\/12.jpg' \/><\/center> $\\dfrac{{{m}^{2}}-3m+9}{{{m}^{3}}-27}-\\dfrac{1}{m-3} $$=\\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{m^3-27}$ ","explain":"<span class='basic_left'> Ta c\u00f3:<br\/> $ \\dfrac{{{m}^{2}}-3m+9}{{{m}^{3}}-27}-\\dfrac{1}{m-3}$<br\/>$=\\dfrac{{{m}^{2}}-3m+9}{\\left( m-3 \\right)\\left( {{m}^{2}}+3m+9 \\right)}-\\dfrac{1}{m-3} $<br\/>$ =\\dfrac{{{m}^{2}}-3m+9}{\\left( m-3 \\right)\\left( {{m}^{2}}+3m+9 \\right)}$$-\\dfrac{1.\\left( {{m}^{2}}+3m+9 \\right)}{\\left( m-3 \\right)\\left( {{m}^{2}}+3m+9 \\right)} $<br\/>$ =\\dfrac{{{m}^{2}}-3m+9-{{m}^{2}}-3m-9}{\\left( m-3 \\right)\\left( {{m}^{2}}+3m+9 \\right)} $<br\/>$ =\\dfrac{-6m}{{{m}^{3}}-27} $ <br\/> <span class='basic_pink'> K\u1ebft qu\u1ea3 c\u1ee7a ph\u00e9p t\u00ednh tr\u00ean l\u00e0 $\\dfrac{-6m}{m^3-27}$. <\/span><\/span> "}]}],"id_ques":82},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng trong ph\u00e9p tr\u1eeb c\u00e1c ph\u00e2n th\u1ee9c sau","title_trans":"","temp":"fill_the_blank","correct":[[["1"],["x"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai13/lv2/img\/11.jpg' \/><\/center> $\\dfrac{3}{2x+6}-\\dfrac{x-6}{2{{x}^{2}}+6x}=\\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$ ","explain":"<span class='basic_left'> Ta c\u00f3:<br\/> $\\begin{align} &\\dfrac{3}{2x+6}-\\dfrac{x-6}{2{{x}^{2}}+6x}\\\\&=\\dfrac{3}{2\\left( x+3 \\right)}-\\dfrac{x-6}{2x\\left( x+3 \\right)} \\\\ & =\\dfrac{3.x}{2x\\left( x+3 \\right)}-\\dfrac{x-6}{2x\\left( x+3 \\right)} \\\\ & =\\dfrac{3x-x+6}{2x\\left( x+3 \\right)} \\\\ & =\\dfrac{2x+6}{2x\\left( x+3 \\right)} \\\\ & =\\dfrac{2\\left( x+3 \\right)}{2x\\left( x+3 \\right)} \\\\ & =\\dfrac{1}{x} \\\\ \\end{align}$ <br\/> <span class='basic_pink'> K\u1ebft qu\u1ea3 c\u1ee7a ph\u00e9p t\u00ednh tr\u00ean l\u00e0 $\\dfrac{1}{x}$. <\/span><\/span> "}]}],"id_ques":83},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng trong ph\u00e9p c\u1ed9ng c\u00e1c ph\u00e2n th\u1ee9c sau","title_trans":"","temp":"fill_the_blank","correct":[[["x-2y","x+2y"],["x+2y","x-2y"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai13/lv2/img\/10.jpg' \/><\/center> $\\dfrac{{{x}^{2}}-2xy}{2{{x}^{2}}+5xy+2{{y}^{2}}}$$+\\dfrac{2xy-4{{y}^{2}}}{2{{x}^{2}}+5xy+2{{y}^{2}}}$$=\\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]{})}{2{{x}^{2}}+5xy+2{{y}^{2}}}$ ","explain":"<span class='basic_left'> Ta c\u00f3:<br\/> $\\begin{align} & \\dfrac{{{x}^{2}}-2xy}{2{{x}^{2}}+5xy+2{{y}^{2}}}+\\dfrac{2xy-4{{y}^{2}}}{2{{x}^{2}}+5xy+2{{y}^{2}}} \\\\ & =\\dfrac{{{x}^{2}}-2xy+2xy-4{{y}^{2}}}{2{{x}^{2}}+5xy+2{{y}^{2}}} \\\\ & =\\dfrac{{{x}^{2}}-4{{y}^{2}}}{2{{x}^{2}}+5xy+2{{y}^{2}}} \\\\ & =\\dfrac{\\left( x-2y \\right)\\left( x+2y \\right)}{2{{x}^{2}}+5xy+2{{y}^{2}}} \\\\ \\end{align}$ <br\/> <span class='basic_pink'> K\u1ebft qu\u1ea3 c\u1ee7a ph\u00e9p t\u00ednh tr\u00ean l\u00e0 $\\dfrac{\\left( x-2y \\right)\\left( x+2y \\right)}{2{{x}^{2}}+5xy+2{{y}^{2}}}$. <\/span><\/span> "}]}],"id_ques":84},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng trong ph\u00e9p c\u1ed9ng c\u00e1c ph\u00e2n th\u1ee9c sau","title_trans":"","temp":"fill_the_blank","correct":[[["x-1"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai13/lv2/img\/9.jpg' \/><\/center> $\\dfrac{2{{x}^{2}}-x}{{{x}^{2}}+x+1}+\\dfrac{{{x}^{3}}-2{{x}^{2}}}{{{x}^{2}}+x+1}$$+\\dfrac{x-1}{{{x}^{2}}+x+1}=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$ ","explain":"<span class='basic_left'> Ta c\u00f3:<br\/> $\\dfrac{2{{x}^{2}}-x}{{{x}^{2}}+x+1}+\\dfrac{{{x}^{3}}-2{{x}^{2}}}{{{x}^{2}}+x+1}$$+\\dfrac{x-1}{{{x}^{2}}+x+1} $<br\/>$ =\\dfrac{2{{x}^{2}}-x+{{x}^{3}}-2{{x}^{2}}+x-1}{{{x}^{2}}+x+1} $<br\/>$ =\\dfrac{{{x}^{3}}-1}{{{x}^{2}}+x+1} $<br\/>$ =\\dfrac{\\left( x-1 \\right)\\left( {{x}^{2}}+x+1 \\right)}{{{x}^{2}}+x+1} $<br\/>$ =x-1 $ <br\/> <span class='basic_pink'> K\u1ebft qu\u1ea3 c\u1ee7a ph\u00e9p t\u00ednh tr\u00ean l\u00e0 $x-1$. <\/span><\/span> "}]}],"id_ques":85},{"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/bai13/lv2/img\/8.jpg' \/><\/center> Quy \u0111\u1ed3ng m\u1eabu th\u1ee9c c\u00e1c ph\u00e2n th\u1ee9c $\\dfrac{x-1}{x^2+2x+1}$; $\\dfrac{2+x}{x+1}$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0:","select":["A. $\\dfrac{x-1}{(x+1)^2}$; $\\dfrac{2+x}{(x+1)^2}$ ","B. $\\dfrac{x-1}{(x+1)^2}$; $\\dfrac{(2+x)(x+1)}{(x+1)^2}$ ","C. $\\dfrac{x^2-1}{x+1}$; $\\dfrac{2+x}{x+1}$"],"hint":" <b> Mu\u1ed1n quy \u0111\u1ed3ng m\u1eabu th\u1ee9c c\u00e1c ph\u00e2n th\u1ee9c:<\/b> <br\/> + T\u00ecm m\u1eabu th\u1ee9c chung <br\/> + X\u00e1c \u0111\u1ecbnh c\u00e1c nh\u00e2n t\u1eed ph\u1ee5: nh\u00e2n t\u1eed ph\u1ee5 l\u00e0 th\u01b0\u01a1ng c\u1ee7a m\u1eabu th\u1ee9c chung v\u1edbi t\u1eebng m\u1eabu th\u1ee9c <br\/> + Nh\u00e2n t\u1eed v\u00e0 m\u1eabu c\u1ee7a m\u1ed7i ph\u00e2n th\u1ee9c v\u1edbi nh\u00e2n t\u1eed ph\u1ee5 c\u1ee7a n\u00f3. ","explain":"<span class='basic_left'> Ta c\u00f3: $\\dfrac{x-1}{x^2+2x+1}=\\dfrac{x-1}{(x+1)^2}$ <br\/> M\u1eabu th\u1ee9c chung: $(x+1)^2$ <br\/> Nh\u00e2n t\u1eed ph\u1ee5 c\u1ee7a m\u1eabu th\u1ee9c $(x+1)^2$ l\u00e0: $1$. Do \u0111\u00f3:<br\/> $\\dfrac{x-1}{x^2+2x+1}=\\dfrac{x-1}{(x+1)^2}$$=\\dfrac{(x-1).1}{(x+1)^2.1}=\\dfrac{x-1}{(x+1)^2}$ <br\/> Nh\u00e2n t\u1eed ph\u1ee5 c\u1ee7a m\u1eabu th\u1ee9c $x+1$ l\u00e0: $x+1$. Do \u0111\u00f3:<br\/> $\\dfrac{2+x}{x+1}=\\dfrac{(2+x)(x+1)}{(x+1)^2}$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span>","column":1}]}],"id_ques":86},{"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/bai13/lv2/img\/5.jpg' \/><\/center> Quy \u0111\u1ed3ng m\u1eabu th\u1ee9c c\u00e1c ph\u00e2n th\u1ee9c $\\dfrac{3}{x^2y^3}$; $\\dfrac{7}{10x^5y^2}$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0:","select":["A. $\\dfrac{30x^3}{10x^5y^3}$; $\\dfrac{7y}{10x^5y^3}$ ","B. $\\dfrac{1}{10x^5y^3}$; $\\dfrac{7}{10x^5y^3}$ ","C. $\\dfrac{3x^3}{10x^5y^3}$; $\\dfrac{7}{10x^5y^3}$"],"explain":"<span class='basic_left'> M\u1eabu th\u1ee9c chung: $10x^5y^3$ <br\/> Nh\u00e2n t\u1eed ph\u1ee5 c\u1ee7a m\u1eabu th\u1ee9c $x^2y^3$ l\u00e0: $10x^3$. Do \u0111\u00f3:<br\/> $\\dfrac{3}{x^2y^3}=\\dfrac{3.10x^3}{x^2y^3.10x^3}=\\dfrac{30x^3}{10x^5y^3}$ <br\/> Nh\u00e2n t\u1eed ph\u1ee5 c\u1ee7a m\u1eabu th\u1ee9c $10x^5y^2$ l\u00e0: $y$. Do \u0111\u00f3:<br\/> $\\dfrac{7}{10x^5y^2}=\\dfrac{7.x}{10x^5y^2.y}=\\dfrac{7y}{10x^5y^3}$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span>","column":3}]}],"id_ques":87},{"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/bai13/lv2/img\/4.jpg' \/><\/center> M\u1eabu th\u1ee9c chung c\u1ee7a hai ph\u00e2n th\u1ee9c $\\dfrac{1}{x^2+x+1}$ v\u00e0 $\\dfrac{2}{x-1}$ l\u00e0:","select":["A. $1-x^3$ ","B. $x^2-1$ ","C. $1-x^2$","D. $x^3-1$"],"explain":"<span class='basic_left'> C\u00e1c m\u1eabu th\u1ee9c kh\u00f4ng ph\u00e2n t\u00edch \u0111\u01b0\u1ee3c th\u00e0nh nh\u00e2n t\u1eed.<br\/> V\u1eady m\u1eabu th\u1ee9c chung l\u00e0: $(x^2+x+1)(x-1)=x^3-1$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span> <br\/> <b> L\u01b0u \u00fd: <\/b> C\u00e1c m\u1eabu th\u1ee9c kh\u00f4ng ph\u00e2n t\u00edch \u0111\u01b0\u1ee3c th\u00e0nh nh\u00e2n t\u1eed th\u00ec m\u1eabu th\u1ee9c chung l\u00e0 t\u00edch c\u1ee7a c\u00e1c m\u1eabu th\u1ee9c","column":2}]}],"id_ques":88},{"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/bai13/lv2/img\/3.jpg' \/><\/center> M\u1eabu th\u1ee9c chung c\u1ee7a hai ph\u00e2n th\u1ee9c $\\dfrac{x-1}{x^2+2x-3}$ v\u00e0 $\\dfrac{-1}{x+3}$ l\u00e0:","select":["A. $x+3$ ","B. $x+1$ ","C. $x-1$","D. $x^2+2x$"],"explain":"<span class='basic_left'> Ta c\u00f3: <br\/> $\\begin{align} \\dfrac{x-1}{{{x}^{2}}+2x-3}&=\\dfrac{x-1}{{{x}^{2}}+3x-x-3} \\\\ & =\\dfrac{x-1}{x\\left( x+3 \\right)-\\left( x+3 \\right)} \\\\ & =\\dfrac{x-1}{\\left( x+3 \\right)\\left( x-1 \\right)} \\\\ & =\\dfrac{1}{x+3} \\\\ \\end{align}$ <br\/> Suy ra m\u1eabu th\u1ee9c chung c\u1ee7a $\\dfrac{1}{x+3}$ v\u00e0 $\\dfrac{-1}{x+3}$ l\u00e0: $x+3$ <br\/> V\u1eady m\u1eabu th\u1ee9c chung c\u1ee7a $\\dfrac{x-1}{x^2+2x-3}$ v\u00e0 $\\dfrac{-1}{x+3}$ l\u00e0: $x+3$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span>","column":2}]}],"id_ques":89},{"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/bai13/lv2/img\/2.jpg' \/><\/center> M\u1eabu th\u1ee9c chung c\u1ee7a hai ph\u00e2n th\u1ee9c $\\dfrac{1}{(x-2)(x+3)}$ v\u00e0 $\\dfrac{-2}{x+3}$ l\u00e0:","select":["A. $(x-2)(x+3)$ ","B. $x-2$ ","C. $x+3$","D. $(x+2)(x-3)$"],"explain":"<span class='basic_left'> $BCNN(1;1)=1$ <br\/> S\u1ed1 m\u0169 cao nh\u1ea5t c\u1ee7a $x-2$ l\u00e0 $1;$ s\u1ed1 m\u0169 cao nh\u1ea5t c\u1ee7a $x+3$ l\u00e0 $1.$ <br\/> V\u1eady m\u1eabu th\u1ee9c chung l\u00e0: $(x-2)(x+3)$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A. <\/span>","column":2}]}],"id_ques":90}],"lesson":{"save":0,"level":2}}