Đáp án+Giải thích các bước giải:
`1)P=\frac{\sqrt{x}+1}{x-1}-\frac{x+2}{x\sqrt{x}-1}-\frac{\sqrt{x}+1}{x+\sqrt{x}+1}(x>=0,x\ne1)`
`=\frac{\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+1)ư}-\frac{x+2}{(\sqrt{x}-1)(x+\sqrt{x}+1)}-\frac{\sqrt{x}+1}{x+\sqrt{x}+1}`
`=\frac{1}{\sqrt{x}-1}-\frac{x+2}{(\sqrt{x}-1)(x+\sqrt{x}+1)}-\frac{\sqrt{x}+1}{x+\sqrt{x}+1}`
`=\frac{x+\sqrt{x}+1-x-2-x+1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}`
`=\frac{-x+\sqrt{x}}{(\sqrt{x}-1)(x+\sqrt{x}+1)}`
`=\frac{-\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}`
`=\frac{-\sqrt{x}}{x+\sqrt{x}+1}`
`2)x=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}`
`=\sqrt{(2+\sqrt{3})^2}+\sqrt{(2-\sqrt{3})^2}`
`=2+\sqrt{3}+2-\sqrt{3}`
`=4`
`->\sqrt{x}=\sqrt{4}=2`
`->P=\frac{-2}{4+2+1}`
`=\frac{-2}{7}`
`->x=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}` thì `P=-2/7`
`3)Q=2/P+\sqrt{x}`
`=2:\frac{-\sqrt{x}}[x+\sqrt{x}+1}+\sqrt{x}`
`=\frac{2x+2\sqrt{x}+2}{-\sqrt{x}}+\sqrt{x}`
`=-2\sqrt{x}-2-\frac{2}{\sqrt{x}}+\sqrt{x}`
`=-\sqrt{x}-2-\frac{2}{\sqrt{x}}`
`=-(\sqrt{x}+\frac{2}{\sqrt{x}})-2`
Áp dụng BĐT Cauchy, ta có:
`\sqrt{x}+\frac{2}{\sqrt{x}}>=2\sqrt{\sqrt{x}.\frac{2}{\sqrt{x}}}=2\sqrt{2}`
`->-(\sqrt{x}+\frac{2}{\sqrt{x}})-2<=-2\sqrt{2}-2`
Vậy `Max_Q=-2\sqrt{2}-2` khi `x=2`
