chứng minh A=1/2√1+1/3√2+1/4√3+...+1/2021√2022<2 Ai giải nhanh và đúng mình vote 5 sao

Đáp án:

chứng minh

Giải thích các bước giải:

Ta đi chứng minh :

$\frac{1}{\sqrt[]{x}(x+1)} ≤ \frac{2}{\sqrt[]{x}} - \frac{2}{\sqrt[]{x+1}}$ với $∀ x > 0$

Ta có : $\frac{1}{\sqrt[]{x}(x+1)} = \frac{\sqrt[]{x}}{x(x+1)}$

⇔ $\frac{1}{\sqrt[]{x}(x+1)} = \sqrt[]{x} . \frac{1}{x(x+1)}$

⇔ $\frac{1}{\sqrt[]{x}(x+1)} = \sqrt[]{x} . ( \frac{1}{x} - \frac{1}{x+1} )$

⇔ $\frac{1}{\sqrt[]{x}(x+1)} = \sqrt[]{x} . ( \frac{1}{\sqrt[]{x}} - \frac{1}{\sqrt[]{x+1}} )( \frac{1}{\sqrt[]{x}} + \frac{1}{\sqrt[]{x+1}} )$

⇔ $\frac{1}{\sqrt[]{x}(x+1)} = ( \frac{1}{\sqrt[]{x}} - \frac{1}{\sqrt[]{x+1}} )( \frac{\sqrt[]{x}}{\sqrt[]{x}} + \frac{\sqrt[]{x}}{\sqrt[]{x+1}} )$

⇔ $\frac{1}{\sqrt[]{x}(x+1)} = ( \frac{1}{\sqrt[]{x}} - \frac{1}{\sqrt[]{x+1}} )( 1 + \frac{\sqrt[]{x}}{\sqrt[]{x+1}} )$

Vì $\frac{\sqrt[]{x}}{\sqrt[]{x+1}} < \frac{\sqrt[]{x+1}}{\sqrt[]{x+1}}$ với $∀ x > 0$

hay $\frac{\sqrt[]{x}}{\sqrt[]{x+1}} < 1$

⇒ $\frac{\sqrt[]{x}}{\sqrt[]{x+1}} + 1 < 2$

⇒ $\frac{1}{\sqrt[]{x}(x+1)} < \frac{2}{\sqrt[]{x}} - \frac{2}{\sqrt[]{x+1}}$ ( đpcm )

Áp dụng : 

$\frac{1}{2\sqrt[]{1}} < \frac{2}{1} - \frac{2}{\sqrt[]{2}}$

$\frac{1}{3\sqrt[]{2}} < \frac{2}{\sqrt[]{2}} - \frac{2}{\sqrt[]{3}}$

$\frac{1}{4\sqrt[]{3}} < \frac{2}{\sqrt[]{3}} - \frac{2}{\sqrt[]{4}}$

$...$

$\frac{1}{2021\sqrt[]{2022}} < \frac{2}{\sqrt[]{2021}} - \frac{2}{\sqrt[]{2022}}$

⇒ $A < \frac{2}{1} - \frac{2}{\sqrt[]{2}} + \frac{2}{\sqrt[]{2}} - \frac{2}{\sqrt[]{3}} + ... + \frac{2}{\sqrt[]{2021}} - \frac{2}{\sqrt[]{2022}}$

⇔ $A < 2 - \frac{2}{\sqrt[]{2022}} < 2$

⇔ $A < 2$ ( đpcm )