Đáp án:
$\begin{array}{l}
11)a)\dfrac{{{x^2} - 5}}{{x + \sqrt 5 }} = \dfrac{{{x^2} - {{\left( {\sqrt 5 } \right)}^2}}}{{x + \sqrt 5 }}\\
= \dfrac{{\left( {x + \sqrt 5 } \right)\left( {x - \sqrt 5 } \right)}}{{x + \sqrt 5 }}\\
= x - \sqrt 5 \\
b)\dfrac{{{x^2} + 2\sqrt 2 + 2}}{{{x^2} - 2}}\\
= \dfrac{{{x^2} - 2 + 2\sqrt 2 + 4}}{{{x^2} - 2}}\\
= 1 + \dfrac{{2\sqrt 2 + 4}}{{{x^2} - 2}}\\
B12)\\
a)\sqrt {{x^2} - 2x + 1} = 3\\
\Rightarrow \sqrt {{{\left( {x - 1} \right)}^2}} = 3\\
\Rightarrow \left[ \begin{array}{l}
x - 1 = 3\\
x - 1 = - 3
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 4\\
x = - 2
\end{array} \right.\\
b)\sqrt {4{x^2} - 10x + 25} + 2x = 5\\
\Rightarrow \sqrt {4{x^2} - 10x + 25} = 5 - 2x\left( {dkxd:x \le \dfrac{5}{2}} \right)\\
\Rightarrow 4{x^2} - 10x + 25 = {\left( {2x - 5} \right)^2}\\
\Rightarrow 4{x^2} - 10x + 25 = 4{x^2} - 20x + 25\\
\Rightarrow 10x = 0\\
\Rightarrow x = 0\left( {tmdk} \right)\\
c)\sqrt {{x^2} + x + 1} = x + 1\left( {dkxd:x \ge - 1} \right)\\
\Rightarrow {x^2} + x + 1 = {x^2} + 2x + 1\\
\Rightarrow x = 0\left( {tmdk} \right)\\
d)\sqrt {x - 3} + \sqrt {3 - x} = 0\\
Dkxd:\left\{ \begin{array}{l}
x \ge 3\\
x \le 3
\end{array} \right. \Rightarrow x = 3
\end{array}$
