Đáp án:
$\begin{array}{l}
a)\left| { - x + 5} \right| - \left| {2{x^2} - 3x - 7} \right| = 0\\
\Rightarrow \left| {x - 5} \right| = \left| {2{x^2} - 3x - 7} \right|\\
\Rightarrow \left[ \begin{array}{l}
x - 5 = 2{x^2} - 3x - 7\\
x - 5 = - 2{x^2} + 3x + 7
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
2{x^2} - 4x - 2 = 0\\
2{x^2} - 2x - 12 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
{x^2} - 2x - 1 = 0\\
{x^2} - x - 6 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \sqrt 2 + 1\\
x = - \sqrt 2 + 1\\
x = 3\\
x = - 2
\end{array} \right.\\
b)\left| {4 - 3x} \right| - 5x - 28 = 0\\
\Rightarrow \left| {4 - 3x} \right| = 5x + 28\left( {dk:x \ge - \dfrac{{28}}{5}} \right)\\
\Rightarrow \left[ \begin{array}{l}
4 - 3x = 5x + 28\\
4 - 3x = - 5x - 28
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
8x = - 24\\
2x = - 32
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = - 3\left( {tm} \right)\\
x = - 16\left( {ktm} \right)
\end{array} \right.\\
Vậy\,x = - 3\\
c)2x - \left| { - {x^2} + 3x - 1} \right| + 7 = 0\\
\Rightarrow \left| {{x^2} - 3x + 1} \right| = 2x + 7\left( {dk:x \ge - \dfrac{7}{2}} \right)\\
\Rightarrow \left[ \begin{array}{l}
{x^2} - 3x + 1 = 2x + 7\\
{x^2} - 3x + 1 = - 2x - 7
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
{x^2} - 5x - 6 = 0\\
{x^2} - x + 8 = 0\left( {vn} \right)
\end{array} \right.\\
\Rightarrow \left( {x - 6} \right)\left( {x + 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 6\\
x = - 1
\end{array} \right.\left( {tm} \right)\\
Vậy\,x = - 1;x = 6\\
d)\left| { - {x^2} - 4x + 6} \right| - \left| {{x^2} + 6x - 12} \right| = 0\\
\Rightarrow \left| {{x^2} + 4x - 6} \right| = \left| {{x^2} + 6x - 12} \right|\\
\Rightarrow \left[ \begin{array}{l}
{x^2} + 4x - 6 = {x^2} + 6x - 12\\
{x^2} + 4x - 6 = - {x^2} - 6x + 12
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
2x = 6\\
2{x^2} + 10x - 18 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 3\\
{x^2} + 5x - 9 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 3\\
x = \dfrac{{ - 5 \pm \sqrt {61} }}{2}
\end{array} \right.\\
g)2\left| { - {x^2} - 4x + 5} \right| - \left| {4{x^2} + 5x - 9} \right| = 0\\
\Rightarrow \left| {2{x^2} + 8x - 10} \right| = \left| {4{x^2} + 5x - 9} \right|\\
\Rightarrow \left[ \begin{array}{l}
2{x^2} + 8x - 10 = 4{x^2} + 5x - 9\\
2{x^2} + 8x - 10 = - 4{x^2} - 5x + 9
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
2{x^2} - 3x + 1 = 0\\
6{x^2} + 13x - 19 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 1\\
x = \dfrac{1}{2}\\
x = \dfrac{{ - 19}}{6}
\end{array} \right.
\end{array}$
