Giải phương trình: \(\dfrac{x-1}{2x -3}=\dfrac{-3x+1}{|x+1|}\) (3)

ĐKXĐ: 

\(\left\{ \begin{array}{l}
2x - 3 \ne 0\\
x + 1 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne \dfrac{3}{2}\\
x \ne - 1
\end{array} \right.\)

+ Xét x > –1, khi đó x + 1 > 0 nên |x + 1| = x + 1.

Khi đó pt (3)

\(\begin{array}{l} \Leftrightarrow \dfrac{{x - 1}}{{2x - 3}} = \dfrac{{ - 3x + 1}}{{x + 1}}\\ \Rightarrow \left( {x - 1} \right)\left( {x + 1} \right) \\= \left( { - 3x + 1} \right)\left( {2x - 3} \right)\\ \Leftrightarrow {x^2} - 1 =  - 6{x^2} + 11x - 3\\ \Leftrightarrow 7{x^2} - 11x + 2 = 0\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{{11 + \sqrt {65} }}{{14}}\left( {TM} \right)\\x = \dfrac{{11 - \sqrt {65} }}{{14}}\left( {TM} \right)\end{array} \right.\end{array}\)

+ Xét x < –1, khi đó x + 1 < 0 nên |x + 1| = –x – 1.

Khi đó pt (3)

\(\begin{array}{l} \Leftrightarrow \dfrac{{x - 1}}{{2x - 3}} = \dfrac{{ - 3x + 1}}{{ - x - 1}}\\ \Rightarrow \left( {x - 1} \right)\left( { - x - 1} \right) \\= \left( { - 3x + 1} \right)\left( {2x - 3} \right)\\ \Leftrightarrow  - {x^2} + 1 =  - 6{x^2} + 11x - 3\\ \Leftrightarrow 5{x^2} - 11x + 4 = 0\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{{11 + \sqrt {41} }}{{10}}\left( {loai} \right)\\x = \dfrac{{11 - \sqrt {41} }}{{10}}\left( {loai} \right)\end{array} \right.\end{array}\)

Vậy phương trình có hai nghiệm là \({x_{1,2}} = \dfrac{{11 \pm \sqrt {65} }}{{14}}\).