$Q=(\frac{1}{\sqrt{a}+1}-\frac{1}{a+\sqrt{a}}):\frac{\sqrt{a}-1}{a+2\sqrt{a}+1}$
ĐKXĐ:
$\sqrt{a}+1\neq0$
$a+\sqrt{a}\neq0$
$a+2\sqrt{a}+1\neq0$
$a>0$
$<=>a>0$
$Q=(\frac{1}{\sqrt{a}+1}-\frac{1}{a+\sqrt{a}}):\frac{\sqrt{a}-1}{a+2\sqrt{a}+1}$
=$(\frac{\sqrt{a}}{a+\sqrt{a}}-\frac{1}{a+\sqrt{a}}).\frac{a+2\sqrt{a}+1}{\sqrt{a}-1}$
=$\frac{\sqrt{a}-1}{a+\sqrt{a}}.\frac{(\sqrt{a}+1)^2}{\sqrt{a}-1}$
=$\frac{(\sqrt{a}+1)^2}{\sqrt{a}(\sqrt{a}+1)}$
=$\frac{\sqrt{a}+1}{\sqrt{a}}>1 ∀a>0$
