Đáp án+Giải thích các bước giải:
`P=\frac{1}{\sqrt{x}+2}-\frac{5}{x-\sqrt{x}-6}-\frac{\sqrt{x}-2}{3-\sqrt{x}}(x>=0,x\ne9)`
`=\frac{1}{\sqrt{x}+2}-\frac{5}{x+2\sqrt{x}-3\sqrt{x}-6}+\frac{\sqrt{x}-2}{\sqrt{x}-3}`
`=\frac{\sqrt{x}-3-5+(\sqrt{x}-2)(\sqrt{x}+2)}{(\sqrt{x}+2)(\sqrt{x}-3)}`
`=\frac{\sqrt{x}-8+x-4}{(\sqrt{x}+2)(\sqrt{x}-3)}`
`=\frac{x+\sqrt{x}-12}{(\sqrt{x}+2)(\sqrt{x}-3)}`
`=\frac{x-3\sqrt{x}+4\sqrt{x}-12}{(\sqrt{x}+2)(\sqrt{x}-3)}`
`=\frac{(\sqrt{x}-3)(\sqrt{x}+4)}{(\sqrt{x}+2)(\sqrt{x}-3)}`
`=\frac{\sqrt{x}+4}{\sqrt{x}+2}`
`b)P=\frac{\sqrt{x}+4}{\sqrt{x}+2}`
`=\frac{\sqrt{x}+2+2}{\sqrt{x}+2}`
`=1+\frac{2}{\sqrt{x}+2}`
Vì `\sqrt{x}+2>=2`
`<=>\frac{2}{\sqrt{x}+2}<=1`
`=>1+\frac{2}{\sqrt{x}+2}<=1+1=2`
Dấu "=" xảy ra khi `\sqrt{x}+2=2<=>x=0`
Vậy `Max_P=2` khi `x=0`
