Đáp án:
Giải thích các bước giải:
`B=(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}):\frac{\sqrt{x}+1}{(\sqrt{x}-1)^2}`
ĐK: `x>0 , x \ne 1`
`B=[\frac{1}{\sqrt{x}(\sqrt{x}-1)}+\frac{\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)}].\frac{(\sqrt{x}-1)^2}{\sqrt{x}+1}`
`B=\frac{1+\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)}.\frac{(\sqrt{x}-1)^2}{\sqrt{x}+1}`
`B=\frac{\sqrt{x}-1}{\sqrt{x}}`
`B=1/3`
`⇔ \frac{\sqrt{x}-1}{\sqrt{x}}=1/3`
`⇔ 3\sqrt{x}-3=\sqrt{x}`
`⇔ 2\sqrt{x}=3`
`⇔ \sqrt{x}=3/2`
`⇔ x=9/4\ (TM)`
Vậy `x=9/4` thì `B =1/3`
