Question

P=\(\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{2}{x-4}\right).\left(\sqrt{x}-1+\dfrac{\sqrt{x}-4}{\sqrt{x}}\right)\)(với x>0 và x≠4)

Chứng minh rằng P=\(\sqrt{x}+3\)

 

Answer 1

Ta có \(P=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)-2}{x-4}.\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}-4}{\sqrt{x}}\)
\(=\dfrac{x+3\sqrt{x}}{x-4}.\dfrac{x-4}{\sqrt{x}}=\sqrt{x}+3\)

Answer 2

Với `x > 0,x \ne 4` có:

`P=([\sqrt{x}+1]/[\sqrt{x}-2]-2/[x-4]).(\sqrt{x}-1+[\sqrt{x}-4]/\sqrt{x})`

`P=[(\sqrt{x}+1)(\sqrt{x}+2)-2]/[(\sqrt{x}-2)(\sqrt{x}+2)].[\sqrt{x}(\sqrt{x}-1)+\sqrt{x}-4]/\sqrt{x}`

`P=[x+2\sqrt{x}+\sqrt{x}+2-2]/[x-4].[x-\sqrt{x}+\sqrt{x}-4]/\sqrt{x}`

`P=[x+3\sqrt{x}]/[x-4].[x-4]/\sqrt{x}`

`P=[\sqrt{x}(\sqrt{x}+3)]/\sqrt{x}=\sqrt{x}+3`

    `=>Đpcm`