Question

Chứng minh rằng:

 

\(\frac{\sqrt{a}}{a+\sqrt{a}}+\frac{\sqrt{a}-1}{2\sqrt{a}}\left(\frac{1}{a-\sqrt{a}}+\frac{1}{a+\sqrt{a}}\right)=\frac{\sqrt{a}}{a}\)\(\frac{\sqrt{a}}{a}\)

 

Answer 1

\(\frac{\sqrt{a}}{a+\sqrt{a}}+\frac{\sqrt{a}-1}{2\sqrt{a}}.\left(\frac{1}{a-\sqrt{a}}+\frac{1}{a+\sqrt{a}}\right)\)

\(=\frac{1}{\sqrt{a}+1}+\frac{\sqrt{a}-1}{2a}.\left(\frac{1}{\sqrt{a}-1}+\frac{1}{\sqrt{a}+1}\right)\)

\(=\frac{1}{\sqrt{a}+1}+\frac{\sqrt{a}-1}{2a}.\frac{2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)

\(=\frac{1}{\sqrt{a}+1}+\frac{1}{\sqrt{a}\left(\sqrt{a}+1\right)}\)

\(=\frac{1}{\sqrt{a}}=\frac{\sqrt{a}}{a}\)