Đáp án:
$\begin{array}{l}
a)Dkxd:x \ne - 1;x \ne \dfrac{1}{2}\\
\dfrac{4}{{x + 1}} + 1 = \dfrac{{2x + 1}}{{2x - 1}}\\
\Leftrightarrow \dfrac{{4 + x + 1}}{{x + 1}} = \dfrac{{2x + 1}}{{2x - 1}}\\
\Leftrightarrow \left( {x + 5} \right).\left( {2x - 1} \right) = \left( {x + 1} \right)\left( {2x + 1} \right)\\
\Leftrightarrow 2{x^2} - x + 10x - 5 = 2{x^2} + x + 2x + 1\\
\Leftrightarrow 9x - 5 = 3x + 1\\
\Leftrightarrow 6x = 6\\
\Leftrightarrow x = 1\left( {tm} \right)\\
Vậy\,x = 1\\
b)Dkxd:x \ne - 2;x \ne 3\\
\dfrac{{2x}}{{x + 2}} - \dfrac{{3x + 2}}{{3 - x}} + 7 = 0\\
\Leftrightarrow \dfrac{{2x\left( {3 - x} \right) - \left( {3x + 2} \right)\left( {x + 2} \right) + 7\left( {x + 2} \right)\left( {3 - x} \right)}}{{\left( {x + 2} \right)\left( {3 - x} \right)}} = 0\\
\Leftrightarrow 6x - 2{x^2} - \left( {3{x^2} + 8x + 4} \right) + 7\left( {3x - {x^2} + 6 - 2x} \right) = 0\\
\Leftrightarrow 6x - 2{x^2} - 3{x^2} - 8x - 4 - 7{x^2} + 7x + 42 = 0\\
\Leftrightarrow - 12{x^2} + 5x + 38 = 0\\
\Leftrightarrow \left( { - 12x - 19} \right)\left( {x - 2} \right) = 0\\
\Leftrightarrow x = - \dfrac{{19}}{{12}};x = 2\left( {tmdk} \right)\\
Vậy\,x = - \dfrac{{19}}{{12}};x = 2
\end{array}$
