Đáp án:
$\begin{array}{l}
b)Dkxd:x \ne 0;x \ne 2\\
\dfrac{{x + 2}}{{x - 2}} - \dfrac{1}{x} = \dfrac{2}{{x\left( {x - 2} \right)}}\\
\Rightarrow \dfrac{{x\left( {x + 2} \right) - \left( {x - 2} \right)}}{{x\left( {x - 2} \right)}} = \dfrac{2}{{x\left( {x - 2} \right)}}\\
\Rightarrow {x^2} + 2x - x + 2 = 2\\
\Rightarrow {x^2} + x = 0\\
\Rightarrow x\left( {x + 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
x + 1 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 0\left( {ktm} \right)\\
x = - 1\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = - 1\\
c)Dkxd:x \ne 2;x \ne - 2\\
\dfrac{{x + 1}}{{x - 2}} + \dfrac{{x - 1}}{{x + 2}} = \dfrac{{2\left( {{x^2} + 2} \right)}}{{{x^2} - 4}}\\
\Rightarrow \dfrac{{\left( {x + 1} \right)\left( {x + 2} \right) + \left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{{2{x^2} + 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
\Rightarrow {x^2} + 3x + 2 + {x^2} - 3x + 2 = 2{x^2} + 4\\
\Rightarrow 2{x^2} + 4 = 2{x^2} + 4\left( {tm} \right)
\end{array}$
Vậy phương trình nghiệm đúng với mọi x#2; x#-2
