Question

Cho A=\(\left(\dfrac{\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{3}{\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\)  với x > 0, x khác 4

a) Rút gọn A

b) Tính A với x = 6-2√5

Answer 1

Answer 2

a) Ta có: \(A=\left(\dfrac{\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{3}{\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\)

\(=\dfrac{\sqrt{x}-4+3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}:\dfrac{x-4-x}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{-4}\)

\(=\dfrac{4\left(\sqrt{x}-1\right)}{-4}=-\sqrt{x}+1\)

b) Thay \(x=6-2\sqrt{5}\) vào A, ta được:

\(A=-\sqrt{6-2\sqrt{5}}+1=-\left(\sqrt{5}-1\right)+1=-\sqrt{5}+1+1=2-\sqrt{5}\)