Đáp án:
$\begin{array}{l}
1)a)Dkxd:a > b > 0\\
Q = \dfrac{a}{{\sqrt {{a^2} - {b^2}} }} - \left( {1 + \dfrac{a}{{\sqrt {{a^2} - {b^2}} }}} \right):\dfrac{b}{{a - \sqrt {{a^2} - {b^2}} }}\\
= \dfrac{a}{{\sqrt {{a^2} - {b^2}} }} - \dfrac{{\sqrt {{a^2} - {b^2}} + a}}{{\sqrt {{a^2} - {b^2}} }}.\dfrac{{a - \sqrt {{a^2} - {b^2}} }}{b}\\
= \dfrac{a}{{\sqrt {{a^2} - {b^2}} }} - \dfrac{{{a^2} - \left( {{a^2} - {b^2}} \right)}}{{\sqrt {{a^2} - {b^2}} }}.\dfrac{1}{b}\\
= \dfrac{a}{{\sqrt {{a^2} - {b^2}} }} - \dfrac{{{b^2}}}{{\sqrt {{a^2} - {b^2}} .b}}\\
= \dfrac{a}{{\sqrt {{a^2} - {b^2}} }} - \dfrac{b}{{\sqrt {{a^2} - {b^2}} }}\\
= \dfrac{{a - b}}{{\sqrt {\left( {a - b} \right)\left( {a + b} \right)} }}\\
= \dfrac{{\sqrt {a - b} }}{{\sqrt {a + b} }}\\
b)a = 3b\\
\Rightarrow Q = \dfrac{{\sqrt {3b - b} }}{{\sqrt {3b + b} }} = \dfrac{{\sqrt {2b} }}{{2\sqrt b }} = \dfrac{{\sqrt 2 }}{2}\\
2)a)Dkxd:x \ge 0;x \ne 4\\
P = \dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} + \dfrac{{2\sqrt x }}{{\sqrt x + 2}} + \dfrac{{2 + 5\sqrt x }}{{4 - x}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x + 2} \right) + 2\sqrt x \left( {\sqrt x - 2} \right) - 2 - 5\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x + 3\sqrt x + 2 + 2x - 4\sqrt x - 2 - 5\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{3x - 6\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{3\sqrt x \left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{3\sqrt x }}{{\sqrt x + 2}}\\
b)P = 2\\
\Rightarrow \dfrac{{3\sqrt x }}{{\sqrt x + 2}} = 2\\
\Rightarrow 3\sqrt x = 2\sqrt x + 4\\
\Rightarrow \sqrt x = 4\\
\Rightarrow x = 16\left( {tmdk} \right)\\
\text{Vậy}\,x = 16
\end{array}$
