Đáp án:
$\begin{array}{l}
c)Dkxd:x\# 1;x\# - 1\\
\dfrac{{x + 1}}{{x - 1}} - \dfrac{{x - 1}}{{x + 1}} = \dfrac{4}{{{x^2} - 1}}\\
\Leftrightarrow \dfrac{{{{\left( {x + 1} \right)}^2} - {{\left( {x - 1} \right)}^2}}}{{\left( {x + 1} \right)\left( {x - 1} \right)}} = \dfrac{4}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
\Leftrightarrow {x^2} + 2x + 1 - {x^2} + 2x - 1 = 4\\
\Leftrightarrow 4x = 4\\
\Leftrightarrow x = 1\left( {ktm} \right)
\end{array}$
Vậy phương trình vô nghiệm
$\begin{array}{l}
d)Dkxd:x\# - 3\\
2x - \dfrac{{2{x^2}}}{{x + 3}} = \dfrac{{4x}}{{x + 3}} + \dfrac{2}{7}\\
\Leftrightarrow \dfrac{{7\left( {x + 3} \right).2x - 2{x^2}.7}}{{7\left( {x + 3} \right)}} = \dfrac{{4x.7 + 2.\left( {x + 3} \right)}}{{7\left( {x + 3} \right)}}\\
\Leftrightarrow 14{x^2} + 42x - 14{x^2} = 28x + 2x + 6\\
\Leftrightarrow 12x = 6\\
\Leftrightarrow x = \dfrac{1}{2}\left( {tmdk} \right)\\
Vay\,x = \dfrac{1}{2}
\end{array}$
