Đáp án+Giải thích các bước giải:
`U=(\frac{3\sqrt{x}+6}{x-4}+\frac{\sqrt{x}}{\sqrt{x}-2}):\frac{x-9}{\sqrt{x}-3}(x>=0,x\ne4,x\ne9)`
`=(\frac{3(\sqrt{x}+2)}{(\sqrt{x}+2)(\sqrt{x}-2)}+\frac{\sqrt{x}}{\sqrt{x}-2}).\frac{\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}`
`=\frac{3+\sqrt{x}}{\sqrt{x}-2}.\frac{1}{\sqrt{x}+3}`
`=\frac{1}{\sqrt{x}-2}`
`c)T=U.V`
`=\frac{1}{\sqrt{x}-2}.\frac{\sqrt{x}-2}{2\sqrt{x}+1}(x>=0,x\ne4,x\ne9)`
`=\frac{1}{2\sqrt{x}+1}`
Ta có: `\sqrt{x}>=0`
`=>2\sqrt{x}>=0`
`<=>2\sqrt{x}+1>=1`
`=>\frac{1}{2\sqrt{x}+1}<=1/1=1`
Dấu `=` xảy ra khi `2\sqrt{x}+1=1<=>x=0`
Vậy `GTLNN=1` khi `x=0`
