Đáp án:
\[x = 1\]
Giải thích các bước giải:
Ta có:
\[\begin{array}{l}
\sqrt {2{x^2} - 4x + 3} + {x^2} = 2\left( {x + 1} \right) - \sqrt {3{x^2} - 6x + 7} \\
\Leftrightarrow \left( {\sqrt {2{x^2} - 4x + 3} - 1} \right) + \left( {{x^2} - 2x + 1} \right) + \left( {\sqrt {3{x^2} - 6x + 7} - 2} \right) = 0\\
\Leftrightarrow \frac{{2{x^2} - 4x + 3 - 1}}{{\sqrt {2{x^2} - 4x + 3} + 1}} + {\left( {x - 1} \right)^2} + \frac{{3{x^2} - 6x + 7 - 4}}{{\sqrt {3{x^2} - 6x + 7} + 2}} = 0\\
\Leftrightarrow \frac{{2\left( {{x^2} - 2x + 1} \right)}}{{\sqrt {2{x^2} - 4x + 3} + 1}} + {\left( {x - 1} \right)^2} + \frac{{3\left( {{x^2} - 2x + 1} \right)}}{{\sqrt {3{x^2} - 6x + 7} + 2}} = 0\\
\Leftrightarrow {\left( {x - 1} \right)^2}\left( {\frac{2}{{\sqrt {2{x^2} - 4x + 3} + 1}} + 1 + \frac{3}{{\sqrt {3{x^2} - 6x + 7} + 2}}} \right) = 0\\
\Rightarrow x = 1
\end{array}\]
