Đáp án:
\(\begin{array}{l}
a,\\
\left[ \begin{array}{l}
x = 1\\
x = \dfrac{1}{5}
\end{array} \right.\\
b,\\
x = \dfrac{1}{4}
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
5x\left( {x - 1} \right) = x - 1\\
\Leftrightarrow 5x\left( {x - 1} \right) - \left( {x - 1} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {5x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
5x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \dfrac{1}{5}
\end{array} \right.\\
b,\\
- 4{\left( {x - 1} \right)^2} + \left( {2x - 1} \right)\left( {2x + 1} \right) = - 3\\
\Leftrightarrow - 4.\left( {{x^2} - 2.x.1 + {1^2}} \right) + \left[ {{{\left( {2x} \right)}^2} - {1^2}} \right] = - 3\\
\Leftrightarrow - 4.\left( {{x^2} - 2x + 1} \right) + \left( {4{x^2} - 1} \right) = - 3\\
\Leftrightarrow - 4{x^2} + 8x - 4 + 4{x^2} - 1 + 3 = 0\\
\Leftrightarrow 8x - 2 = 0\\
\Leftrightarrow x = \dfrac{1}{4}
\end{array}\)
