`A=(x-\sqrt{x}+1)/(\sqrt{x})`
Ta có: `x=17-12\sqrt{2}=17-2.3.\sqrt{8}=8-2.3.\sqrt{8}+9=(2\sqrt{2}-3)^2`
`-> \sqrt{x}=\sqrt{(2\sqrt{2}-3)^2}=3-2\sqrt{2}`
Thay `x=17-12\sqrt{2}; \sqrt{x}=3-2\sqrt{2}` vào A ta có:
`A=(17-12\sqrt{2}-3+2\sqrt{2}+1)/(3-2\sqrt{2})`
`A=(15-10\sqrt{2})/(3-2\sqrt{2})`
`A=(5(3-2\sqrt{2}))/(3-2\sqrt{2})`
`A=5`
Vậy `A=5` khi `x=17-12\sqrt{2}`