Đáp án:
$\begin{array}{l}
Đặt:{x^2} - 2x + 2 = u\\
\Rightarrow {x^2} - 2x + 1 + 1 = u\\
\Rightarrow {\left( {x - 1} \right)^2} + 1 = u \ge 1\\
Pt:\frac{1}{u} + \frac{1}{{u + 1}} = \frac{9}{{2\left( {u + 2} \right)}}\\
\Rightarrow \frac{{u + 1 + u}}{{u\left( {u + 1} \right)}} = \frac{9}{{2u + 4}}\\
\Rightarrow \left( {2u + 1} \right)\left( {2u + 4} \right) = 9u\left( {u + 1} \right)\\
\Rightarrow 4{u^2} + 8u + 2u + 4 = 9{u^2} + 9u\\
\Rightarrow 5{u^2} - u - 4 = 0\\
\Rightarrow 5{u^2} - 5u + 4u - 4 = 0\\
\Rightarrow \left( {u - 1} \right)\left( {5u + 4} \right) = 0\\
\Rightarrow u = 1\left( {do:u \ge 1} \right)\\
\Rightarrow {x^2} - 2x + 2 = 1\\
\Rightarrow {x^2} - 2x + 1 = 0\\
\Rightarrow x = 1
\end{array}$
Vậy x=1
