Question

Cho biêu thức A=\(\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

a. RG A

Answer 1

A=\(\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
`=((x-2+\sqrtx)/(x+2\sqrtx).({\sqrt{x}+1}{\sqrt{x}-1})`
`=((\sqrtx-1)(\sqrtx+2))/(\sqrtx(\sqrtx+2)).({\sqrt{x}+1}{\sqrt{x}-1})`
`=(\sqrtx+1)/\sqrtx`

Answer 2

A=\(\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
`=((x-2+\sqrtx)/(x+2\sqrtx).({\sqrt{x}+1}/{\sqrt{x}-1})`
`=((\sqrtx-1)(\sqrtx+2))/(\sqrtx(\sqrtx+2)).({\sqrt{x}+1}/{\sqrt{x}-1})`
`=(\sqrtx+1)/\sqrtx`

Answer 3

a) Ta có: \(A=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\left(\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\dfrac{x+2\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)