Đáp án:
d) \(\left[ \begin{array}{l}
x = - 2\\
x = 4
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a)7 - 2x - 4 = - x - 4\\
\to x = 7\\
b){\left( {x - 5} \right)^2} - 3\left( {x - 5} \right) = 0\\
\to \left( {x - 5} \right)\left( {x - 5 - 3} \right) = 0\\
\to \left[ \begin{array}{l}
x - 5 = 0\\
x - 8 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 5\\
x = 8
\end{array} \right.\\
c)DK:x \ne \pm 2\\
\dfrac{{\left( {x - 1} \right)\left( {x - 2} \right) - x\left( {x + 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = - \dfrac{{7x - 6}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
\to \dfrac{{{x^2} - 3x + 2 - {x^2} - 2x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = - \dfrac{{7x - 6}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
\to - 5x + 2 = - 7x + 6\\
\to 2x = 4\\
\to x = 2\left( l \right)\\
\to x \in \emptyset \\
d)\left| {5 - 2x} \right| = 7 - x\\
\to \left[ \begin{array}{l}
5 - 2x = 7 - x\left( {DK:\dfrac{5}{2} \ge x} \right)\\
5 - 2x = - 7 + x\left( {DK:\dfrac{5}{2} < x} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 2\\
3x = 12
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 2\\
x = 4
\end{array} \right.\left( {TM} \right)
\end{array}\)
