a) $\frac{2x+1}{2x^{2}-3x}-\frac{2}{2x-3}=\frac{1}{x}$ b) $\frac{1}{x-6}-\frac{2}{6+x}=\frac{3x+6}{x^{2}-36}$

Đáp án:

$\begin{array}{l}
a)Dkxd:x \ne 0;x \ne \dfrac{3}{2}\\
\dfrac{{2x + 1}}{{2{x^2} - 3x}} - \dfrac{2}{{2x - 3}} = \dfrac{1}{x}\\
 \Leftrightarrow \dfrac{{2x + 1 - 2x}}{{x\left( {2x - 3} \right)}} = \dfrac{{2x - 3}}{{x\left( {2x - 3} \right)}}\\
 \Leftrightarrow 1 = 2x - 3\\
 \Leftrightarrow 2x = 4\\
 \Leftrightarrow x = 2\left( {tm} \right)\\
Vậy\,x = 2\\
b)Dk:x \ne 6;x \ne  - 6\\
\dfrac{1}{{x - 6}} - \dfrac{2}{{6 + x}} = \dfrac{{3x + 6}}{{{x^2} - 36}}\\
 \Leftrightarrow \dfrac{{x + 6 - 2\left( {x - 6} \right)}}{{\left( {x + 6} \right)\left( {x - 6} \right)}} = \dfrac{{3x + 6}}{{\left( {x + 6} \right)\left( {x - 6} \right)}}\\
 \Leftrightarrow x + 6 - 2x + 12 = 3x + 6\\
 \Leftrightarrow  - x + 12 = 3x\\
 \Leftrightarrow 4x = 12\\
 \Leftrightarrow x = 3\left( {tm} \right)\\
Vậy\,x = 3
\end{array}$