Question

\(\dfrac{3\sqrt{x}}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{2-\sqrt{x}}+\dfrac{8\sqrt{x}}{x-4}\)

Answer 1

Với `x >= 0,x \ne 4` có:

   `[3\sqrt{x}]/[\sqrt{x}+2]+\sqrt{x}/[2-\sqrt{x}]+[8\sqrt{x}]/[x-4]`

`=[3\sqrt{x}(\sqrt{x}-2)-\sqrt{x}(\sqrt{x}+2)+8\sqrt{x}]/[(\sqrt{x}+2)(\sqrt{x}-2)]`

`=[3x-6\sqrt{x}-x-2\sqrt{x}+8\sqrt{x}]/[(\sqrt{x}+2)(\sqrt{x}-2)]`

`=[2x]/[x-4]`

Answer 2

\(=\dfrac{3\sqrt{x}\left(\sqrt{x-2}\right)-\sqrt{x}\left(\sqrt{x+2}\right)+8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{3x-6\sqrt{x}-x-2\sqrt{x}+8\sqrt{2}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{2x}{x-4}\)