Đáp án:`x>0`.
`1)x=64(tmđk)`
`=>A=(2+sqrt{64})/64`
`=10/64=5/32`
Vậy `A=5/32` khi `x=64`
`2)B=(sqrtx-1)/sqrtx+(2sqrtx+1)/(x+sqrtx)`
`B=((sqrtx-1)(sqrtx+1))/(sqrtx(sqrtx+1))+(2sqrtx+1)/(sqrtx(sqrtx+1))`
`B=(x-1+2sqrtx+1)/(sqrtx(sqrtx+1))`
`B=(x+2sqrtx)/(sqrtx(sqrtx+1))`
`B=(sqrtx(sqrtx+2))/(sqrtx(sqrtx+1))`
`B=(sqrtx+2)/(sqrtx+1)`
`3)A/B>3/2`
`<=>(sqrtx+2)/x:(sqrtx+2)/(sqrtx+1)>3/2`
`<=>(sqrtx+1)/x>3/2`
`<=>(2sqrtx+2)/x>3`
`=>2sqrtx+2>3x`
`<=>x-2/3sqrtx-2/3<0`
`<=>(sqrtx-1/3)^2<2/3+1/9=7/9`
`<=>sqrtx-1/3<sqrt7/3`
`<=>x<(sqrt7+1)^2/3`
Vậy `0<=x<(sqrt7+1)^2/3` thì `A/B>3/2.`
