Đáp án:

\(\begin{array}{l}
b)x = 1\\
c)\left[ \begin{array}{l}
x = 2\\
x =  - 1
\end{array} \right.\\
d)\left[ \begin{array}{l}
x =  - 1 - \sqrt 3 \\
x = \dfrac{{ - 1 - \sqrt 7 }}{2}
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
b)\left| {2{x^2} - 5x + 3} \right| =  - 2{x^2} + 2\\
 \to \left[ \begin{array}{l}
2{x^2} - 5x + 3 =  - 2{x^2} + 2\left( {DK:2{x^2} - 5x + 3 \ge 0} \right)\\
2{x^2} - 5x + 3 = 2{x^2} - 2\left( {DK:2{x^2} - 5x + 3 < 0} \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
4{x^2} - 5x + 1 = 0\left( {DK:\left[ \begin{array}{l}
x \ge \dfrac{3}{2}\\
x \le 1
\end{array} \right.} \right)\\
 - 5x =  - 5\left( {DK:1 < x < \dfrac{3}{2}} \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 1\\
x = \dfrac{1}{4}\left( l \right)\\
x = 1\left( l \right)
\end{array} \right.\\
 \to x = 1\\
c)\left| {3{x^2} - 7x + 2} \right| =  - {x^2} + 5x - 6\\
 \to \left[ \begin{array}{l}
3{x^2} - 7x + 2 =  - {x^2} + 5x - 6\left( {DK:3{x^2} - 7x + 2 \ge 0} \right)\\
3{x^2} - 7x + 2 = {x^2} - 5x + 6\left( {DK:3{x^2} - 7x + 2 < 0} \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
4{x^2} - 12x + 8 = 0\left( {DK:\left[ \begin{array}{l}
x \ge 2\\
x \le \dfrac{1}{3}
\end{array} \right.} \right)\\
2{x^2} - 2x - 4 = 0\left( {DK:\dfrac{1}{3} < x < 2} \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 2\\
x = 1\left( l \right)\\
x = 2\left( l \right)\\
x =  - 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 2\\
x =  - 1
\end{array} \right.\\
d)\left| {3{x^2} + 4x - 5} \right| = {x^2} - 1\\
 \to \left[ \begin{array}{l}
3{x^2} + 4x - 5 = {x^2} - 1\left( {DK:3{x^2} + 4x - 5 \ge 0} \right)\\
3{x^2} + 4x - 5 =  - {x^2} + 1\left( {DK:3{x^2} + 4x - 5 < 0} \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
2{x^2} + 4x - 4 = 0\left( {DK:\left[ \begin{array}{l}
x \ge \dfrac{{ - 2 + \sqrt {19} }}{3}\\
x \le \dfrac{{ - 2 - \sqrt {19} }}{3}
\end{array} \right.} \right)\\
4{x^2} + 4x - 6 = 0\left( {DK:\dfrac{{ - 2 - \sqrt {19} }}{3} < x < \dfrac{{ - 2 + \sqrt {19} }}{3}} \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x =  - 1 + \sqrt 3 \left( l \right)\\
x =  - 1 - \sqrt 3 \\
x = \dfrac{{ - 1 + \sqrt 7 }}{2}\left( l \right)\\
x = \dfrac{{ - 1 - \sqrt 7 }}{2}
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x =  - 1 - \sqrt 3 \\
x = \dfrac{{ - 1 - \sqrt 7 }}{2}
\end{array} \right.
\end{array}\)