Đáp án:
$\begin{array}{l}
B3)\\
b)Dkxd:x\# 0;x\# 2;x\# - 2\\
A = \dfrac{1}{{2 - x}} = 2 - x\\
\Leftrightarrow {\left( {2 - x} \right)^2} = 1\\
\Leftrightarrow \left[ \begin{array}{l}
2 - x = 1\\
2 - x = - 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 3
\end{array} \right.\left( {tmdk} \right)\\
Vậy\,x = 1;x = 3\\
B4)\\
b)Dkxd:x\# 0;x\# - 1\\
A = \dfrac{{x - 1}}{3} = \dfrac{{1 - {x^2}}}{{25}}\\
\Leftrightarrow 25\left( {x - 1} \right) = 3\left( {1 - {x^2}} \right)\\
\Leftrightarrow 25\left( {x - 1} \right) + 3\left( {x - 1} \right)\left( {x + 1} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {25 + 3x + 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\left( {tm} \right)\\
x = - \dfrac{{28}}{3}\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = \dfrac{{ - 28}}{3};x = 1
\end{array}$
