Toán Lớp 9: Rút gọn
$\frac{1}{1 + \sqrt{2}}$ + $\frac{1}{\sqrt{2} + \sqrt{3}}$ + … + $\frac{1}{\sqrt{99} + \sqrt{100}}$
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Toán Lớp 9: Rút gọn
$\frac{1}{1 + \sqrt{2}}$ + $\frac{1}{\sqrt{2} + \sqrt{3}}$ + … + $\frac{1}{\sqrt{99} + \sqrt{100}}$
Comments ( 2 )
\quad \dfrac{1}{1 + \sqrt2} + \dfrac{1}{\sqrt 2+ \sqrt3} + \cdots + \dfrac{1}{\sqrt{99} + \sqrt{100}}\\
= \dfrac{\sqrt2 – 1}{\left(1 + \sqrt2\right)\left(\sqrt2 – 1\right)} +\dfrac{\sqrt3 – \sqrt2}{\left(\sqrt2 + \sqrt3\right)\left(\sqrt3 – \sqrt2\right)} +\cdots + \dfrac{\sqrt{100} – \sqrt{99}}{\left(\sqrt{99} + \sqrt{100}\right)\left(\sqrt{100} – \sqrt{99}\right)} \\
= \dfrac{\sqrt2 – 1}{2 – 1} + \dfrac{\sqrt3 – \sqrt2}{3 – 2} + \cdots + \dfrac{\sqrt{100} – \sqrt{99}}{100 – 99}\\
= \sqrt2 – 1 + \sqrt3 – \sqrt2 + \cdots + \sqrt{100} – \sqrt{99}\\
= \sqrt{100} – 1\\
= 10 – 1\\
= 9
\end{array}\)