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a/\(\dfrac{8}{x-8}+1+\dfrac{11}{x-11}+1=\dfrac{9}{x-9}+1+\dfrac{10}{x-10}+1\)
=>\(\dfrac{8+x-8}{x-8}+\dfrac{11+x-11}{x-11}=\dfrac{9+x-9}{x-9}+\dfrac{10+x-10}{x-10}\)
=>\(\dfrac{x}{x-8}+\dfrac{x}{x-11}-\dfrac{x}{x-9}-\dfrac{x}{x-10}=0\)
=>x.\(\left(\dfrac{1}{x-8}+\dfrac{1}{x-11}+\dfrac{1}{x-9}+\dfrac{1}{x-10}\right)=0\)
=>x=0
b/\(\dfrac{x}{x-3}-1+\dfrac{x}{x-5}-1=\dfrac{x}{x-4}-1+\dfrac{x}{x-6}-1\)
=>\(\dfrac{x-x+3}{x-3}+\dfrac{x-x+5}{x-5}-\dfrac{x-x+4}{x-4}-\dfrac{x-6+6}{x-6}=0\)
=>\(\dfrac{3}{x-3}+\dfrac{5}{x-5}-\dfrac{4}{x-4}-\dfrac{6}{x-6}=0\)
Đến đây thì bạn giải giống câu a
a) \(\dfrac{x+1}{2}+\dfrac{3x-2}{3}=\dfrac{x-7}{12}\)
\(\Leftrightarrow\dfrac{6\left(x+1\right)+4\left(3x-2\right)}{12}=\dfrac{x-7}{12}\)
\(\Leftrightarrow6\left(x+1\right)+4\left(3x-2\right)=x-7\)
\(\Leftrightarrow6x+6+12x-8=x-7\)
\(\Leftrightarrow6x+12x-x=-7-6+8\)
\(\Leftrightarrow17x=-5\)
\(\Leftrightarrow x=\dfrac{-5}{17}\)
Vậy .........................
b) \(\dfrac{2x}{x-3}-\dfrac{5}{x+3}=\dfrac{x^2+21}{x^2-9}\left(ĐKXĐ:x\ne\pm3\right)\)
\(\Leftrightarrow\dfrac{2x\left(x+3\right)-5\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+21}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow2x\left(x+3\right)-5\left(x-3\right)=x^2+21\)
\(\Leftrightarrow2x^2+6x-5x+15=x^2+21\)
\(\Leftrightarrow2x^2-x^2+x+15-21=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow x^2-2x+3x-6=0\)
\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(n\right)\\x=-3\left(l\right)\end{matrix}\right.\)
Vậy \(S=\left\{2\right\}\)
d) \(\left(x-4\right)\left(7x-3\right)-x^2+16=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3\right)-\left(x^2-16\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3\right)-\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3-x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(6x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\6x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{6}\end{matrix}\right.\)
Vậy .........................
P/s: các câu còn lại tương tự, bn tự giải nha
a)\(x-\dfrac{5x+2}{6}=\dfrac{7-3x}{4}\)
\(\Leftrightarrow\dfrac{12x-10x-4}{12}=\dfrac{21-9x}{12}\)
\(\Leftrightarrow2x-4=21-9x\)
\(\Leftrightarrow2x-4-21+9x=0\)
\(\Leftrightarrow11x-25=0\)
\(\Leftrightarrow x=\dfrac{25}{11}\)
b)\(\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\)
\(\Leftrightarrow\dfrac{30x+9}{36}=\dfrac{36+24+32x}{36}\)
\(\Leftrightarrow30x+9=60+32x\)
\(\Leftrightarrow30x+9-60-32x=0\)
\(\Leftrightarrow-2x-51=0\)
\(\Leftrightarrow x=-\dfrac{51}{2}\)
c)\(\dfrac{x}{3}-\dfrac{2x+1}{2}=\dfrac{x}{6}-6\)
\(\Leftrightarrow\dfrac{2x-6x-3}{6}=\dfrac{x-36}{6}\)
\(\Leftrightarrow-4x-3=x-36\)
\(\Leftrightarrow-4x-3-x+36=0\)
\(\Leftrightarrow-5x+33=0\)
\(\Leftrightarrow x=\dfrac{33}{5}\)
d)\(\dfrac{2+x}{3}-\dfrac{1}{2}x=\dfrac{1-2x}{4}+\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{8+4x-6x}{12}=\dfrac{3-6x+3}{12}\)
\(\Leftrightarrow8-2x=6-6x\)
\(\Leftrightarrow8-2x-6+6x=0\)
\(\Leftrightarrow4x+2=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Tính lại xem đúng không nha 
a) \(x-\dfrac{5x+2}{6}=\dfrac{7-3x}{4}\)
\(\Leftrightarrow\dfrac{24x}{24}-\dfrac{4\left(5x+2\right)}{24}=\dfrac{6\left(7-3x\right)}{24}\)
\(\Leftrightarrow24x-4\left(5x+2\right)=6\left(7-3x\right)\)
\(\Leftrightarrow24x-20x-8=42-18x\)
\(\Leftrightarrow4x-8=42-18x\)
\(\Leftrightarrow4x+18x=42+8\)
\(\Leftrightarrow22x=50\)
\(\Leftrightarrow x=\dfrac{25}{11}\)
Vậy S\(=\left\{\dfrac{25}{11}\right\}\)
a,\(x-\frac{5x+2}{6}=\frac{7-3x}{4}\)
=> \(\frac{12x}{12}-\frac{\left(5x+2\right)2}{12}=\frac{\left(7-3x\right)3}{12}\)
=>\(\frac{12x-10x-4}{12}=\frac{21-9x}{12}\)
=>(khử mẫu)
=>\(12x-10x-4=21-9x\)
=>11x=25
=>x=25/11
b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)
=>30x+9=36+32x+24
=>32x+60=30x+9
=>2x=-51
=>x=-51/2
c: \(\Leftrightarrow2x-3\left(2x+1\right)=x+6x\)
=>7x=2x-6x-3
=>7x=-4x-3
=>11x=-3
=>x=-3/11
d: \(\Leftrightarrow4\left(x+2\right)-6x=3\left(1-2x+1\right)\)
=>4x+8-6x=3(-2x+2)
=>-2x+8+6x-6=0
=>4x+2=0
=>x=-1/2
b: Đặt \(x^2-6x-2=a\)
Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)
=>(a+2)(a+7)=0
\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)
=>x(x-6)(x-1)(x-5)=0
hay \(x\in\left\{0;1;6;5\right\}\)
c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)
\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)
\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)
\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)
=>26x=-3
hay x=-3/26
6). ĐK x\(\ne\)...
PT \(\Leftrightarrow\dfrac{4}{2x^3+3x^2-8x-12}-\dfrac{1\left(2x+3\right)}{2x^3+3x^2-8x-12}-\dfrac{4\left(x-2\right)}{2x^3+3x^2-8x-12}+\dfrac{1\left(x^2-4\right)}{2x^3+3x^2-8x-12}=0\)(cái này bạn lấy 2x3+3x2-8x-12 chia cho các mẫu khác bằng phương pháp chia hoocner)
\(\Leftrightarrow4-\left(2x+3\right)-4\left(x-2\right)+\left(x^2-4\right)=0\\ \Leftrightarrow x^2-6x+5=0\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
\(a,\dfrac{3\left(5x-2\right)}{4}-2=\dfrac{7x}{3}-5\left(x-7\right)\)
\(\Leftrightarrow\dfrac{15x-6-8}{4}=\dfrac{7x-15\left(x-7\right)}{3}\)
\(\Leftrightarrow\dfrac{15x-14}{4}=\dfrac{7x-15x+105}{3}\)
\(\Leftrightarrow\dfrac{45x-42}{12}=\dfrac{-32x+420}{12}\)
\(\Leftrightarrow45x+32x=420+42\)
\(\Leftrightarrow77x=462\)
\(\Leftrightarrow x=6\)
\(b,\dfrac{x+5}{2}+\dfrac{3-2x}{4}=x-\dfrac{7+x}{6}\)
\(\Leftrightarrow\dfrac{2x+10+3-2x}{4}=\dfrac{6x-7-x}{6}\)
\(\Leftrightarrow\dfrac{13}{4}=\dfrac{5x-7}{6}\)
\(\Leftrightarrow2\left(5x-7\right)=3.13\)
\(\Leftrightarrow10x-14=39\)
\(\Leftrightarrow10x=53\)
\(\Leftrightarrow x=5,3\)
\(c,\dfrac{x-3}{11}+\dfrac{x+1}{3}=\dfrac{x+7}{9}-1\)
\(\Leftrightarrow\dfrac{3x-9+11x+11}{33}=\dfrac{x+7-9}{9}\)
\(\Leftrightarrow\dfrac{14x+2}{33}=\dfrac{x-2}{9}\)
\(\Leftrightarrow33\left(x-2\right)=9\left(14x+2\right)\)
\(\Leftrightarrow33x-66=126x+18\)
\(\Leftrightarrow-93x=84\)
\(\Leftrightarrow x=-\dfrac{28}{31}\)
\(d,\dfrac{3x-0,4}{2}+\dfrac{1,5-2x}{3}=\dfrac{x+0,5}{5}\)
\(\Leftrightarrow\dfrac{3\left(3x-0,4\right)+2\left(1,5-2x\right)}{6}=\dfrac{x+0,5}{5}\)
\(\Leftrightarrow\dfrac{9x-1,2+3-4x}{6}=\dfrac{x+0,5}{5}\)
\(\Leftrightarrow\dfrac{5x+1,8}{6}=\dfrac{x+0,5}{5}\)
\(\Leftrightarrow5\left(5x+1,8\right)=6\left(x+0,5\right)\)
\(\Leftrightarrow25x+9=6x+3\)
\(\Leftrightarrow19x=-6\)
\(\Leftrightarrow x=-\dfrac{6}{19}\)
\(\Leftrightarrow77x=378\)
\(\Leftrightarrow x=\dfrac{54}{11}\)
a: \(\Leftrightarrow x^2-4-x+4=-2\left(x^2-6x+8\right)\)
=>-2x^2+12x-16=x^2-x
=>-3x^2+13x-16=0
=>3x^2-13x+16=0
Δ=(-13)^2-4*3*16=169-192<0
=>PTVN
b: \(\Leftrightarrow\left(x-2\right)\left(x-3\right)+\left(x+1\right)\left(x+3\right)=2x^2+6\)
=>x^2-5x+6+x^2+4x+3=2x^2+6
=>-x+9=6
=>-x=-3
=>x=3(loại)
c: \(\Leftrightarrow4\left(2x-1\right)-3\left(x-6\right)=6\left(3x-2\right)\)
=>8x-4-3x+12=18x-12
=>-5x+8=18x-12
=>-23x=-20
=>x=20/23
d: \(\Leftrightarrow5\left(3x-1\right)+8x-21=12\left(x+2\right)-40\)
=>15x-5+8x-21=12x+24-40
=>23x-26=12x-16
=>11x=10
=>x=10/11