Đáp án+Giải thích các bước giải:
`a)A=\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\frac{2a+\sqrt{a}}{\sqrt{a}}+1(a>0)`
`=\frac{\sqrt{a}(a\sqrt{a}+1)}{a-\sqrt{a}+1}-\frac{\sqrt{a}(2\sqrt{a}+1)}{\sqrt{a}}+1`
`=\frac{\sqrt{a}(\sqrt{a}+1)(a-\sqrt{a}+1)}{a-\sqrt{a}+1}-2\sqrt{a}-1+1`
`=\sqrt{a}(\sqrt{a}+1)-2\sqrt{a}`
`=a+\sqrt{a}-2\sqrt{a}`
`=a-\sqrt{a}`
`b)A=a-\sqrt{a}`
`=a-2.\sqrt{a}.1/2 +1/4 -1/4`
`=(\sqrt{a}-1/2)^2-1/4`
Ta có: `(\sqrt{a}-1/2)^2>=0`
`=>(\sqrt{a}-1/2)^2-1/4>=-1/4`
Dấu "=" xảy ra khi `\sqrt{a}=1/2<=>a=1/4`
Vậy `GTNNNN=-1/4` khi `a=1/4`
