`1)B=\frac{x+7}{\sqrt{x}}`
Thế `x=25` vào `B`,ta có:
`\frac{25+7}{\sqrt{25}}`
`=\frac{32}{5}`
Vậy `B=32/5` khi `x=25`
`2)A=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}-1}{\sqrt{x}-3}-\frac{2x-\sqrt{x}-3}{x-9}`
`=\frac{\sqrt{x}(\sqrt{x}-3)+(2\sqrt{x}-1)(\sqrt{x}+3)-2x+\sqrt{x}+3}{(\sqrt{x}+3)(\sqrt{x}-3)}`
`=\frac{x-3\sqrt{x}+2x+6\sqrt{x}-\sqrt{x}-3-2x+\sqrt{x}+3}{x-9}`
`=\frac{x+3\sqrt{x}}{x-9}`
`=\frac{\sqrt{x}(\sqrt{x}+3)}{x-9}`
`=\frac{\sqrt{x}}{\sqrt{x}-3}`
`3)S=1/A+B`
`->S=1:\frac{\sqrt{x}}{\sqrt{x}-3}+\frac{x+7}{\sqrt{x}}`
`=\frac{\sqrt{x}-3}{\sqrt{x}}+\frac{x+7}{\sqrt{x}}`
`=\frac{\sqrt{x}-3+7+x}{\sqrt{x}}`
`=\frac{x+\sqrt{x}+4}{\sqrt{x}}`
`=\sqrt{x}+1+\frac{4}{\sqrt{x}}`
Áp dụng `BĐT` cô-si
`->\sqrt{x}+\frac{4}{\sqrt{x}}>=2\sqrt{\sqrt{x}.\frac{4}{\sqrt{x}}=2\sqrt{4}=4`
`=>\sqrt{x}+\frac{4}{\sqrt{x}}>=4`
`<=>\sqrt{x}+1+\frac{4}{\sqrt{x}}>=5`
Vậy `GTNNNN=5` khi `x=4`
