Đáp án:
b) x=-1
Giải thích các bước giải:
\(\begin{array}{l}
C5:\\
a)2\left| {{x^2} + 3x - 4} \right| = x - 1\\
\to \left[ \begin{array}{l}
2\left( {{x^2} + 3x - 4} \right) = x - 1\left( {DK:x \ge 1} \right)\\
2\left( {{x^2} + 3x - 4} \right) = - x + 1\left( {DK:x < 1} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
2{x^2} + 5x - 7 = 0\\
2{x^2} + 7x - 9 = 0
\end{array} \right. \to \left[ \begin{array}{l}
x = 1\left( {TM} \right)\\
x = - \dfrac{7}{2}\left( l \right)\\
x = - \dfrac{9}{2}\left( {TM} \right)
\end{array} \right.\\
b)\sqrt {{x^2} - 3x + 5} = 1 - 2x\\
\to {x^2} - 3x + 5 = 1 - 4x + 4{x^2}\left( {DK:\dfrac{1}{2} \ge x} \right)\\
\to 3{x^2} - x - 4 = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{4}{3}\left( l \right)\\
x = - 1
\end{array} \right.
\end{array}\)
