`A=x-sqrtx+1` ĐK : `x>=0`
`=(sqrtx)^2-2 . sqrtx . 1/2+1/4+1-1/4`
`=(sqrtx-1/2)^2+3/4>=3/4`
Dấu "=" xảy ra khi : `(sqrtx-1/2)^2=0`
`<=> sqrtx=1/2`
`<=> x=1/4 \ \ (tm)`
Vậy `A_(min)=3/4 <=> x=1/4`
$$$$
`B=4x-2sqrtx-1` ĐK : `x>=0`
`=(2sqrtx)^2-2.2sqrtx . 1/2+1/4-1-1/4`
`=(2sqrtx-1/2)^2-5/4>=-5/4`
Dấu "=" xảy ra khi : `(2sqrtx-1/2)^2=0`
`<=> 2sqrtx-1/2=0`
`<=> 2sqrtx=1/2`
`<=> sqrtx=1/4`
`<=> x=1/16 \ \ (tm)`
Vậy `B_(min)=-5/4 <=> x=1/16`