Tính

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thaotran19

Ta có : $\dfrac{1}{1+\sqrt{2}}=\dfrac{1-\sqrt{2}}{1-2}=-1+\sqrt{2}$
$\dfrac{1}{\sqrt{2}+\sqrt{3}}=\dfrac{\sqrt{2}-\sqrt{3}}{2-3}=-\sqrt{2}+\sqrt{3}$
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$\dfrac{1}{\sqrt{99}+\sqrt{100}}=\dfrac{\sqrt{99}-\sqrt{100}}{99-100}=-\sqrt{99}+\sqrt{100}$
$=>A=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-.....-\sqrt{99}+\sqrt{100}=9$
 
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