Đáp án:
a. \(\left[ \begin{array}{l}
x = 1\\
x = - 3
\end{array} \right.\)
b. \(x = \frac{2}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.{(x - 1)^2} + 2(1 - {x^2}) = 0\\
\to {(x - 1)^2} + 2\left( {1 - x} \right)\left( {1 + x} \right) = 0\\
\to {(x - 1)^2} - 2\left( {x - 1} \right)\left( {1 + x} \right) = 0\\
\to \left( {x - 1} \right)\left( {x - 1 - 2 - 2x} \right) = 0\\
\to \left[ \begin{array}{l}
x - 1 = 0\\
- x - 3 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = - 3
\end{array} \right.\\
b.DK:x \ne \left\{ {0;2} \right\}\\
\frac{{3x}}{{2x - 4}} - \frac{1}{x} = \frac{2}{{{x^2} - 2x}}\\
\to \frac{{3x}}{{2\left( {x - 2} \right)}} - \frac{1}{x} = \frac{2}{{x\left( {x - 2} \right)}}\\
\to \frac{{3{x^2} - 2x + 4}}{{2x\left( {x - 2} \right)}} = \frac{4}{{2x\left( {x - 2} \right)}}\\
\to 3{x^2} - 2x = 0\\
\to \left[ \begin{array}{l}
x = 0\left( l \right)\\
x = \frac{2}{3}\left( {TM} \right)
\end{array} \right.
\end{array}\)
