Đáp án+Giải thích các bước giải:
a,
`A=(\sqrt{x}-\frac{x+2}{\sqrt{x}+1}):(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-4}{1-x})(x≥0;x\ne1)`
`A=(\frac{x+\sqrt{x}-x-2}{\sqrt{x}+1}):(\frac{x-\sqrt{x}+\sqrt{x}-4}{(\sqrt{x}+1)(\sqrt{x}-1)})`
`A=\frac{\sqrt{x}-2}{\sqrt{x}+1}:\frac{x-4}{(\sqrt{x}+1)(\sqrt{x}-1)}`
`A=\frac{\sqrt{x}-2}{\sqrt{x}+1}.\frac{(\sqrt{x}+1)(\sqrt{x}-1)}{(\sqrt{x}-2)(\sqrt{x}+2)}`
`A=\frac{\sqrt{x}-1}{\sqrt{x}+2}`
b,
`x=4-2\sqrt{3}`
`⇒x=(\sqrt{3}-1)^2`
Thay vào `A` ta được:
`A=\frac{\sqrt{(\sqrt{3}-1)^2}-1}{\sqrt{(\sqrt{3}-1)^2}+2}`
`A=\frac{\sqrt{3}-1-1}{\sqrt{3}-1+2}`
`A=\frac{\sqrt{3}-2}{\sqrt{3}+1}`
`A=\frac{5-3\sqrt{3}}{2}`
c,
`A<\frac{1}{3}`
`⇔\frac{\sqrt{x}-1}{\sqrt{x}+2}<\frac{1}{3}`
`⇔\frac{\sqrt{x}-1}{\sqrt{x}+2}-\frac{1}{3}<0`
`⇔\frac{3\sqrt{x}-3-\sqrt{x}-2}{3\sqrt{x}+6}<0`
`⇔\frac{2\sqrt{x}-5}{3\sqrt{x}+6}<0`
`3\sqrt{x}+6≥6>0`
`⇒2\sqrt{x}-5<0`
`⇔2\sqrt{x}<5`
`⇔\sqrt{x}<\frac{5}{2}`
`⇔x<\frac{25}{4}`
Vậy `0≤x≤\frac{25}{4};x\ne1` thì `A<\frac{1}{3}`
