Câu hỏi của Nguyễn Tiến Dũng

Question

Chưng minh rằng:$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+.....+\frac{1}{\sqrt{100}}>10$

Lightning Farron · Accepted Answer

Ta có:

$\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}=\frac{1}{10}$

$\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}=\frac{1}{10}$

$\frac{1}{\sqrt{3}}>\frac{1}{\sqrt{100}}=\frac{1}{10}$

$......................$

$\frac{1}{\sqrt{99}}>\frac{1}{\sqrt{100}}=\frac{1}{10}$

Cộng theo vế ta có: $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{99}}>\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}=\frac{99}{10}$

$\Rightarrow\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{99}}+\frac{1}{\sqrt{100}}>\frac{99}{10}+\frac{1}{10}=\frac{100}{10}=10$

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