1)$\forall a>0$ Ta có :
$[a\sqrt{a+1}+(a+1).\sqrt{a}].(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{a+1}})=\sqrt{a(a+1)}-a+a+1+\sqrt{a(a+1})=1$
$\Rightarrow \color{blue}{ \frac{1}{a\sqrt{a+1}+(a+1).\sqrt{a}}=\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{a+1}}}$
$a=1\Rightarrow \frac{1}{1 \sqrt{2}+2\sqrt{1}}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}$
$a=2 \Rightarrow \frac{1}{2 \sqrt{3}+3\sqrt{2}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}$
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$a= 960 \Rightarrow \frac{1}{960 \sqrt{961}+961\sqrt{960}}=\frac{1}{\sqrt{960}}-\frac{1}{\sqrt{961}}$
$\Rightarrow VT=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{961}}=\color{red}{1-\frac{1}{\sqrt{961}}}$