Đáp án:

\(\left[ \begin{array}{l}
x = \dfrac{{10}}{3}\\
x = 0\\
x =  - \dfrac{1}{2}
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
{\left( {{x^2} + 2x - 5} \right)^2} = {\left( {{x^2} - x + 5} \right)^2}\\
 \to \left| {{x^2} + 2x - 5} \right| = \left| {{x^2} - x + 5} \right|\\
 \to \left[ \begin{array}{l}
{x^2} + 2x - 5 = {x^2} - x + 5\\
{x^2} + 2x - 5 =  - {x^2} + x - 5
\end{array} \right.\\
 \to \left[ \begin{array}{l}
3x = 10\\
2{x^2} + x = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{{10}}{3}\\
x\left( {2x + 1} \right) = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{{10}}{3}\\
x = 0\\
x =  - \dfrac{1}{2}
\end{array} \right.
\end{array}\)

$\begin{array}{l} {\left( {{x^2} + 2x - 5} \right)^2} = {\left( {{x^2} - x + 5} \right)^2}\\ \to \left| {{x^2} + 2x - 5} \right| = \left| {{x^2} - x + 5} \right|\\ \to \left[ \begin{array}{l} {x^2} + 2x - 5 = {x^2} - x + 5\\ {x^2} + 2x - 5 = - {x^2} + x - 5 \end{array} \right.\\ \to \left[ \begin{array}{l} 3x = 10\\ 2{x^2} + x = 0 \end{array} \right.\\ \to \left[ \begin{array}{l} x = \dfrac{{10}}{3}\\ x\left( {2x + 1} \right) = 0 \end{array} \right.\\ \to \left[ \begin{array}{l} x = \dfrac{{10}}{3}\\ x = 0\\ x = - \dfrac{1}{2} \end{array} \right. \end{array}$