Đáp án:
\[x = 3\]
Giải thích các bước giải:
\(\begin{array}{l}
DKXD:\,\,\,x \ge 2\\
\sqrt {x + 1} + \sqrt {x - 2} + 2\sqrt {{x^2} - x - 2} = 13 - 2x\\
\Leftrightarrow \sqrt {x + 1} + \sqrt {x - 2} + 2\sqrt {{x^2} - x - 2} + 2x - 13 = 0\\
\Leftrightarrow \left( {\sqrt {x + 1} - 2} \right) + \left( {\sqrt {x - 2} - 1} \right) + 2.\left( {\sqrt {{x^2} - x - 2} - 2} \right) + \left( {2x - 6} \right) = 0\\
\Leftrightarrow \dfrac{{\left( {\sqrt {x + 1} - 2} \right)\left( {\sqrt {x + 1} + 2} \right)}}{{\sqrt {x + 1} + 2}} + \dfrac{{\left( {\sqrt {x - 2} - 1} \right)\left( {\sqrt x - 2} \right) + 1}}{{\sqrt {x - 2} + 1}}\\
+ 2.\dfrac{{\left( {\sqrt {{x^2} - x - 2} - 2} \right)\left( {\sqrt {{x^2} - x - 2} + 2} \right)}}{{\sqrt {{x^2} - x - 2} + 2}} + 2.\left( {x - 3} \right) = 0\\
\Leftrightarrow \dfrac{{{{\sqrt {x + 1} }^2} - {2^2}}}{{\sqrt {x + 1} + 2}} + \dfrac{{{{\sqrt {x - 2} }^2} - {1^2}}}{{\sqrt {x - 2} + 1}} + 2.\dfrac{{{{\sqrt {{x^2} - x - 2} }^2} - {2^2}}}{{\sqrt {{x^2} - x - 2} + 2}} + 2.\left( {x - 3} \right) = 0\\
\Leftrightarrow \dfrac{{x + 1 - 4}}{{\sqrt {x + 1} + 2}} + \dfrac{{x - 2 - 1}}{{\sqrt {x - 2} + 1}} + 2.\dfrac{{{x^2} - x - 2 - 4}}{{\sqrt {{x^2} - x - 2} + 2}} + 2.\left( {x - 3} \right) = 0\\
\Leftrightarrow \dfrac{{x - 3}}{{\sqrt {x + 1} + 2}} + \dfrac{{x - 3}}{{\sqrt {x - 2} + 1}} + 2.\dfrac{{{x^2} - x - 6}}{{\sqrt {{x^2} - x - 2} + 2}} + 2.\left( {x - 3} \right) = 0\\
\Leftrightarrow \dfrac{{x - 3}}{{\sqrt {x + 1} + 2}} + \dfrac{{x - 3}}{{\sqrt {x - 2} + 1}} + 2.\dfrac{{\left( {x - 3} \right)\left( {x + 2} \right)}}{{\sqrt {{x^2} - x - 2} + 2}} + 2.\left( {x - 3} \right) = 0\\
\Leftrightarrow \left( {x - 3} \right).\left[ {\dfrac{1}{{\sqrt {x + 1} + 2}} + \dfrac{1}{{\sqrt {x - 2} + 1}} + 2.\dfrac{{x + 2}}{{\sqrt {{x^2} - x - 2} + 2}} + 2} \right] = 0\\
x \ge 2 \Rightarrow \dfrac{1}{{\sqrt {x + 1} + 2}} + \dfrac{1}{{\sqrt {x - 2} + 1}} + 2.\dfrac{{x + 2}}{{\sqrt {{x^2} - x - 2} + 2}} + 2 > 0\\
\Rightarrow x - 3 = 0\\
\Leftrightarrow x = 3
\end{array}\)
