Câu 10.
a) Điều kiện: `x\ge0,x\ne1`
`P=(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}):\frac{\sqrt{x}}{x-2\sqrt{x}+1}`
`P=(\frac{1}{\sqrt{x}(\sqrt{x}-1)}+\frac{1}{\sqrt{x}-1}):\frac{\sqrt{x}}{(\sqrt{x}-1)^2}`
`P=\frac{1+\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)}.\frac{(\sqrt{x}-1)^2}{\sqrt{x}}`
`P=\frac{(1+\sqrt{x}).(\sqrt{x}-1)}{x}`
`P=\frac{x-1}{x}`
b) Ta có: `P>\frac{1}{2}`
`⇔``\frac{x-1}{x}>\frac{1}{2}`
`⇔` `(x-1).2>x`
`⇔` `2x-2-x>0`
`⇔` `x-2>0`
`⇔` `x>2`
Vậy: với `x>2` thì `P>\frac{1}{2}`