Question

rút gọn

\(\left(\dfrac{1}{\sqrt{a}-\sqrt{a-b}}+\dfrac{1}{\sqrt{a}+\sqrt{a+b}}\right):\left(1+\dfrac{\sqrt{a+b}}{\sqrt{a-b}}\right)\)

Answer 1

đkxđ: \(\sqrt{a}\ne\sqrt{a-b}\) ; a>0; b>0;a≠b

Rút gọn:

Đặt: \(A=\left(\dfrac{1}{\sqrt{a}-\sqrt{a-b}}+\dfrac{1}{\sqrt{a}+\sqrt{a+b}}\right):\left(1+\dfrac{\sqrt{a+b}}{\sqrt{a-b}}\right)\)

\(=\left(\dfrac{\sqrt{a}+\sqrt{a-b}}{a-a+b}+\dfrac{\sqrt{a}-\sqrt{a+b}}{a-a-b}\right):\left(\dfrac{\sqrt{a-b}+\sqrt{a+b}}{\sqrt{a-b}}\right)\)

\(=\dfrac{\sqrt{a}+\sqrt{a-b}-\sqrt{a}+\sqrt{a+b}}{b}\cdot\dfrac{\sqrt{a-b}}{\sqrt{a-b}+\sqrt{a+b}}=\dfrac{\sqrt{a+b}+\sqrt{a-b}}{b}\cdot\dfrac{\sqrt{a-b}}{\sqrt{a-b}+\sqrt{a+b}}=\dfrac{\sqrt{a-b}}{b}\)