## Tính: $\frac{1}{2√1+1√2}$ + $\frac{1}{3√2+2√3}$ +….+ $\frac{1}{100√99+99√100}$

Question

Tính: $\frac{1}{2√1+1√2}$ + $\frac{1}{3√2+2√3}$ +….+ $\frac{1}{100√99+99√100}$

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4 tháng 2021-08-02T11:59:24+00:00 1 Answers 10 views 0

1. Đáp án:

$\dfrac{9}{10}$

Giải thích các bước giải:

$\dfrac{1}{2\sqrt1 + 1\sqrt2} + \dfrac{1}{3\sqrt2 + 2\sqrt3} + \cdots + \dfrac{1}{100\sqrt{99} + 99\sqrt{100}}$

$= \dfrac{1}{\sqrt1.\sqrt2.(\sqrt2 + \sqrt1)} + \dfrac{1}{\sqrt2.\sqrt3.(\sqrt3 + \sqrt2)} + \cdots+ \dfrac{1}{\sqrt{99}.\sqrt{100}.(\sqrt{100} + \sqrt{99})}$

$= \dfrac{\sqrt2 – \sqrt1}{\sqrt1.\sqrt2.(\sqrt2 + \sqrt1)(\sqrt2 – \sqrt1)} + \dfrac{\sqrt3 – \sqrt2}{\sqrt2.\sqrt3.(\sqrt3 + \sqrt2)(\sqrt3 – \sqrt2)} + \cdots+ \dfrac{\sqrt{100} – \sqrt{99}}{\sqrt{99}.\sqrt{100}.(\sqrt{100} + \sqrt{99})(\sqrt{100} – \sqrt{99})}$

$= \dfrac{\sqrt2 – \sqrt1}{\sqrt1.\sqrt2} + \dfrac{\sqrt3 – \sqrt2}{\sqrt2.\sqrt3} + \cdots+ \dfrac{\sqrt{100} – \sqrt{99}}{\sqrt{99}.\sqrt{100}}$

$= \dfrac{1}{\sqrt1} – \dfrac{1}{\sqrt2} + \dfrac{1}{\sqrt2} – \dfrac{1}{\sqrt3} + \cdots + \dfrac{1}{\sqrt{99}} – \dfrac{1}{\sqrt{100}}$

$= 1 – \dfrac{1}{10}$

$= \dfrac{9}{10}$