Đáp án:
$\begin{array}{l}
2)Dkxd:x \ne 2;x \ne - 2\\
\dfrac{1}{{x - 2}} + \dfrac{1}{{x + 2}} = \dfrac{{{x^2} - x + 2}}{{{x^2} - 4}}\\
\Rightarrow \dfrac{{x + 2 + x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{{{x^2} - x + 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
\Rightarrow 2x = {x^2} - x + 2\\
\Rightarrow {x^2} - 3x + 2 = 0\\
\Rightarrow \left( {x - 2} \right)\left( {x + 1} \right) = 0\\
\Rightarrow x = - 1\left( {do:x \ne 2} \right)\\
3)Dkxd:x \ne 3;x \ne - 3\\
\dfrac{{2x}}{{x - 3}} = \dfrac{{{x^2} + 11x - 6}}{{{x^2} - 9}}\\
\Rightarrow 2x\left( {x + 3} \right) = {x^2} + 11x - 6\\
\Rightarrow 2{x^2} + 6x = {x^2} + 11x - 6\\
\Rightarrow {x^2} - 5x + 6 = 0\\
\Rightarrow \left( {x - 3} \right)\left( {x - 2} \right) = 0\\
\Rightarrow x = 2\left( {do:x \ne 3} \right)\\
4)3{x^2} + \left( {1 - \sqrt 3 } \right).x + \sqrt 3 - 4 = 0\\
\Rightarrow 3{x^2} - 3x + \left( {4 - \sqrt 3 } \right).x + \sqrt 3 - 4 = 0\\
\Rightarrow \left( {x - 1} \right).\left( {3x + 4 - \sqrt 3 } \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 1\\
x = \dfrac{{\sqrt 3 - 4}}{3}
\end{array} \right.\\
6)x - 2\sqrt x = 6 - 3\sqrt x \\
\Rightarrow x + \sqrt x - 6 = 0\\
\Rightarrow \left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right) = 0\\
\Rightarrow \sqrt x = 2\\
\Rightarrow x = 4\\
7)y - 2\sqrt y = 36 + 3\sqrt y \\
\Rightarrow y - 5\sqrt y - 36 = 0\\
\Rightarrow \left( {\sqrt y - 9} \right)\left( {\sqrt y + 4} \right) = 0\\
\Rightarrow \sqrt y = 9\\
\Rightarrow y = 81\\
8)\left( {\sqrt 2 + 1} \right).{x^2} - \left( {3 + 2\sqrt 2 } \right).x + \sqrt 2 + 2 = 0\\
\Rightarrow \left( {\sqrt 2 + 1} \right).{x^2} - \left( {\sqrt 2 + 1} \right).x - \left( {\sqrt 2 + 2} \right).x\\
+ \sqrt 2 + 2 = 0\\
\Rightarrow \left( {x - 1} \right)\left( {\left( {\sqrt 2 + 1} \right).x - \sqrt 2 - 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 1\\
x = \sqrt 2
\end{array} \right.
\end{array}$
