Question

rút gọn biểu thức

D=\(\dfrac{x\sqrt{2}}{2\sqrt{x}+x\sqrt{2}}\)+\(\dfrac{\sqrt{2x}-2}{x-2}\)

Answer 1

\(D=\dfrac{x\sqrt{2}}{2\sqrt{x}+x\sqrt{2}}+\dfrac{\sqrt{2x}-2}{x-2}\)

\(ĐK:\left\{{}\begin{matrix}x\ge0\\x\ne2\end{matrix}\right.\)

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

\(D=\dfrac{\sqrt{x}}{\left(\sqrt{2}+\sqrt{x}\right)}+\dfrac{\sqrt{2}}{\sqrt{x}+\sqrt{2}}\)

\(D=\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{2}+\sqrt{x}}\)

\(D=1\)

Answer 2

Với `x >= 0,x \ne 2` có:

`D=[x\sqrt{2}]/[2\sqrt{x}+x\sqrt{2}]+[\sqrt{2x}-2]/[x-2]`

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

`D=\sqrt{x}/[\sqrt{2}+\sqrt{x}]+\sqrt{2}/[\sqrt{x}+\sqrt{2}]`

`D=[\sqrt{x}+\sqrt{2}]/[\sqrt{x}+\sqrt{2}]=1`