Đáp án: x=3
Giải thích các bước giải:
$\begin{array}{l}
Dkxd: - 5 \le x \le 5\\
f\left( x \right) = {x^2} - 3x + 2\\
\Rightarrow f'\left( x \right) = 2x - 3\\
\Rightarrow f''\left( x \right) = 2\\
4f'\left( x \right) - \left( {2x - 5} \right)f''\left( x \right) - x + 1 = 2\left( {\sqrt {25 - {x^2}} } \right)\\
\Rightarrow 4.\left( {2x - 3} \right) - \left( {2x - 5} \right).2 - x + 1 = 2\sqrt {25 - {x^2}} \\
\Rightarrow 3x - 1 = 2\sqrt {25 - {x^2}} \left( {dk:x \ge \frac{1}{3}} \right)\\
\Rightarrow 9{x^2} - 6x + 1 = 4\left( {25 - {x^2}} \right)\\
\Rightarrow 13{x^2} - 6x - 99 = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 3\left( {tm} \right)\\
x = - \frac{{33}}{{13}}\left( {ktm} \right)
\end{array} \right.\\
Vậy\,x = 3
\end{array}$
