Question

giải ci tiết hộ e vs ạ

rút gọn

\(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right)\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)vs x>0

Answer 1

Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

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

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}+1}=1\)

Answer 2

Với `x > 0` có:

`B=(x/[x+3\sqrt{x}]+1/[\sqrt{x}+3])(1-2/\sqrt{x}+6/[x+3\sqrt{x}])`

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

`B=[\sqrt{x}(\sqrt{x}+1)]/[\sqrt{x}(\sqrt{x}+3)] . [x+3\sqrt{x}-2\sqrt{x}-6+6]/[\sqrt{x}(\sqrt{x}+3)]`

`B=[\sqrt{x}+1]/[\sqrt{x}+3] . [x+\sqrt{x}]/[\sqrt{x}(\sqrt{x}+3)]`

`B=[\sqrt{x}+1]/[\sqrt{x}+3] . [\sqrt{x}+1]/[\sqrt{x}+3]`

`B=[(\sqrt{x}+1)^2]/[(\sqrt{x}+3)^2]`