Đáp án+Giải thích các bước giải:
`x=3-2\sqrt{2}`
`=>\sqrt{x}=\sqrt{3-2\sqrt{2}}=\sqrt{(\sqrt{2}-1)^2}=|\sqrt{2}-1|=\sqrt{2}-1`
`->Q=\frac{\sqrt{2}-1-1}{\sqrt{2}-1+1}`
`=\frac{\sqrt{2}-2}{\sqrt{2}}`
`=\frac{\sqrt{2}(1-\sqrt{2})}{\sqrt{2}}`
`=1-\sqrt{2}`
Vậy `Q=1-\sqrt{2}` khi `x=3-2\sqrt{2}`
`A=P.Q`
`=>A=(\frac{x\sqrt{x}-1}{x-\sqrt{x}}+\frac{x\sqrt{x}+1}{x+\sqrt{x}}-\frac{4}{\sqrt{x}}).\frac{\sqrt{x}-1}{\sqrt{x}+1}(x>0,x\ne1)`
`=(\frac{\sqrt{x}^3-1^3}{\sqrt{x}(\sqrt{x}-1)}+\frac{\sqrt{x}^3+1^3}{\sqrt{x}(\sqrt{x}+1)}-\frac{4}{\sqrt{x}}).\frac{\sqrt{x}-1}{\sqrt{x}+1}`
`=(\frac{(\sqrt{x}-1)(x+\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)}+\frac{(\sqrt{x}+1)(x-\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}+1)}-\frac{4}{\sqrt{x}}).\frac{\sqrt{x}-1}{\sqrt{x}+1}`
`=\frac{x+\sqrt{x}+1+x-\sqrt{x}+1-4}{\sqrt{x}}.\frac{\sqrt{x}-1}{\sqrt{x}+1}`
`=\frac{2x-2}{\sqrt{x}}.\frac{\sqrt{x}-1}{\sqrt{x}+1}`
`=\frac{2(sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}}.\frac{\sqrt{x}-1}{\sqrt{x}+1}`
`=\frac{2(\sqrt{x}-1)(\sqrt{x}-1)}{\sqrt{x}}`
`=\frac{2(x-2\sqrt{x}+1)}{\sqrt{x}}`
`=\frac{2x-4\sqrt{x}+2}{\sqrt{x}}`
