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T

thuyduong1851998

Ta có
[TEX]VP=\frac{1}{\sqrt{n+1}+\sqrt{n}}=\frac{1(\sqrt{n+1}-\sqrt{n})}{(\sqrt{n+1}+\sqrt{n})(\sqrt{n+1}-\sqrt{n})}= \frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\sqrt{n+1}-\sqrt{n}= VT (dpcm)[/TEX]

Vậy $\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}$
 
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vy000

$\sqrt[]{n+1}-\sqrt[]{n}=\dfrac 1{\sqrt[]{n+1}+\sqrt[]{n}}$

$\Leftrightarrow (\sqrt[]{n+1}-\sqrt[]{n})(\sqrt[]{n+1}+\sqrt[]{n})=1$

$\Leftrightarrow n+1-n=1$

$\Leftrightarrow 1=1$

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