Trả lời:
a, \(\frac{2}{2x+1}-\frac{3}{2x-1}=\frac{4}{4x^2-1}\)\(\left(đkxđ:x\ne\pm\frac{1}{2}\right)\)
\(\Leftrightarrow\frac{2\left(2x-1\right)-3\left(2x+1\right)}{4x^2-1}=\frac{4}{4x^2-1}\)
\(\Rightarrow4x-2-6x-3=4\)
\(\Leftrightarrow-2x-5=4\)
\(\Leftrightarrow-2x=9\)
\(\Leftrightarrow x=\frac{-9}{2}\left(tm\right)\)
Vậy \(S=\left\{\frac{-9}{2}\right\}\)
b, \(\frac{2x}{x-1}+\frac{18}{x^2+2x-3}=\frac{2x-5}{x+3}\)\(\left(đkxđ:x\ne1;x\ne-3\right)\)
\(\Leftrightarrow\frac{2x\left(x+3\right)+18}{x^2+2x-3}=\frac{\left(2x-5\right)\left(x-1\right)}{x^2+2x-3}\)
\(\Rightarrow2x^2+6x+18=2x^2-7x+5\)
\(\Leftrightarrow2x^2+6x-2x^2+7x=5-18\)
\(\Leftrightarrow13x=-13\)
\(\Leftrightarrow x=-1\)\(\left(tm\right)\)
Vậy \(S=\left\{-1\right\}\)
c, \(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)\(\left(đkxđ:x\ne1\right)\)
\(\Leftrightarrow\frac{x^2+x+1+2x^2-5}{x^3-1}=\frac{4\left(x-1\right)}{x^3-x}\)
\(\Rightarrow3x^2+x-4=4x-4\)
\(\Leftrightarrow3x^2+x-4x=-4+4\)
\(\Leftrightarrow3x^2-3x=0\)
\(\Leftrightarrow3x\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x=1\left(ktm\right)\end{cases}}}\)
Vậy \(S=\left\{0\right\}\)
d, \(\frac{11}{x}=\frac{9}{x+1}+\frac{2}{x-4}\)\(\left(đkxđ:x\ne0;x\ne-1;x\ne4\right)\)
\(\Leftrightarrow\frac{11\left(x+1\right)\left(x-4\right)}{x\left(x+1\right)\left(x-4\right)}=\frac{9x\left(x-4\right)+2x\left(x+1\right)}{x\left(x+1\right)\left(x-4\right)}\)
\(\Rightarrow11\left(x^2-3x-4\right)=9x^2-36x+2x^2+2x\)
\(\Leftrightarrow11x^2-33x-44=11x^2-34x\)
\(\Leftrightarrow11x^2-33x-11x^2+34x=44\)
\(\Leftrightarrow x=44\)\(\left(tm\right)\)
Vậy \(S=\left\{44\right\}\)
e, \(\frac{14}{3x-12}-\frac{2+x}{x-4}=\frac{3}{8-2x}-\frac{5}{6}\)\(\left(đkxđ:x\ne4\right)\)
\(\Leftrightarrow\frac{14}{3\left(x-4\right)}-\frac{2+x}{x-4}=\frac{-3}{2\left(x-4\right)}-\frac{5}{6}\)
\(\Leftrightarrow\frac{28-6\left(2+x\right)}{6\left(x-4\right)}=\frac{-9-5\left(x-4\right)}{6\left(x-4\right)}\)
\(\Rightarrow28-12-6x=-9-5x+20\)
\(\Leftrightarrow16-6x=11-5x\)
\(\Leftrightarrow-6x+5x=11-16\)
\(\Leftrightarrow-x=-5\)
\(\Leftrightarrow x=5\)\(\left(tm\right)\)
Vậy \(S=\left\{5\right\}\)
a)\(\frac{2}{2x+1}\)\(-\frac{3}{2x-1}\)\(=\frac{4}{4x^2-1}\)
<=> \(\frac{2\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}\)\(-\frac{3\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}\)\(=\frac{4}{\left(2x-1\right)\left(2x+1\right)}\)
<=>\(\frac{4x-2-6x-3}{\left(2x-1\right)\left(2x+1\right)}\)\(=\frac{4}{\left(2x-1\right)\left(2x+1\right)}\)
=> -2x-5=4
<=>-2x=9
<=>\(x=\frac{-9}{2}\)
c/ 
d/ 
)(x-
) = 0 g/ (3x-1)(2x-3)(x+5) = 0 
h)
m)
= 8 – x k) 
= – 4x +7
q) 
