Question

Cho biểu thức M= \(\dfrac{a}{\sqrt{ab}+b}+\dfrac{b}{\sqrt{ab}-a}-\dfrac{a+b}{\sqrt{ab}}\)nhau.với hai số a, b dương khác

a/ Rút gọn M

b/Tính giá trị của M khi a=\(\sqrt{6+2\sqrt{5}}\),b=\(\sqrt{6-2\sqrt{5}}\)

Answer 1

a: ta có: \(M=\dfrac{a}{\sqrt{ab}+b}+\dfrac{b}{\sqrt{ab}-a}-\dfrac{a+b}{\sqrt{ab}}\)

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

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

\(=\dfrac{-\sqrt{ab}\left(a+b\right)+\left(a-b\right)\left(a+b\right)}{\sqrt{ab}\left(a-b\right)}-\dfrac{a^2-b^2}{\sqrt{ab}\left(a-b\right)}\)

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

\(=-\dfrac{1}{a-b}\)

Answer 2

b: Thay \(a=\sqrt{5}+1\) và \(b=\sqrt{5}-1\) vào M, ta được:

\(M=\dfrac{-1}{\sqrt{5}+1-\sqrt{5}+1}=\dfrac{-1}{2}\)