Câu hỏi của Nguyễn Minh Anh

Question

Rút gọn biểu thức sau :

$A= \dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{5}} +...+\dfrac{1}{\sqrt{23}+\sqrt{25}} $

Minh Nguyễn Cao · Accepted Answer

$A=\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+...+\frac{1}{\sqrt{23}+\sqrt{25}}$

$2A=\frac{2}{\sqrt{3}+\sqrt{1}}+\frac{2}{\sqrt{5}+\sqrt{3}}+...+\frac{2}{\sqrt{25}+\sqrt{23}}$$2A=\frac{2\left(\sqrt{3}-\sqrt{1}\right)}{\left(\sqrt{3}+\sqrt{1}\right)\left(\sqrt{3}-\sqrt{1}\right)}+\frac{2\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+...+\frac{2\left(\sqrt{25}-\sqrt{23}\right)}{\left(\sqrt{25}+\sqrt{23}\right)\left(\sqrt{25}-\sqrt{23}\right)}$

$2A=\frac{2\left(\sqrt{3}-\sqrt{1}\right)}{2}+\frac{2\left(\sqrt{5}-\sqrt{3}\right)}{2}+...+\frac{2\left(\sqrt{25}-\sqrt{23}\right)}{2}$

$2A=\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+\sqrt{25}-\sqrt{23}$

$2A=\sqrt{25}-\sqrt{1}$

$2A=4$

$A=2$

Câu trả lời:

$A=\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+...+\frac{1}{\sqrt{23}+\sqrt{25}}$

$2A=\frac{2}{\sqrt{3}+\sqrt{1}}+\frac{2}{\sqrt{5}+\sqrt{3}}+...+\frac{2}{\sqrt{25}+\sqrt{23}}$$2A=\frac{2\left(\sqrt{3}-\sqrt{1}\right)}{\left(\sqrt{3}+\sqrt{1}\right)\left(\sqrt{3}-\sqrt{1}\right)}+\frac{2\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+...+\frac{2\left(\sqrt{25}-\sqrt{23}\right)}{\left(\sqrt{25}+\sqrt{23}\right)\left(\sqrt{25}-\sqrt{23}\right)}$

$2A=\frac{2\left(\sqrt{3}-\sqrt{1}\right)}{2}+\frac{2\left(\sqrt{5}-\sqrt{3}\right)}{2}+...+\frac{2\left(\sqrt{25}-\sqrt{23}\right)}{2}$

$2A=\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+\sqrt{25}-\sqrt{23}$

$2A=\sqrt{25}-\sqrt{1}$

$2A=4$

$A=2$

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