Đáp án:
$a)min=\dfrac{7}{4}\Leftrightarrow x=\dfrac{1}{4}\\ b)min=1005\sqrt{2} \Leftrightarrow x=1006$
Giải thích các bước giải:
$a)x-\sqrt{x}+2\\ =x-\sqrt{x}+\dfrac{1}{4}+\dfrac{7}{4}\\ =\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{7}{4} \ge \dfrac{7}{4} \ \forall \ x$
Dấu "=" xảy ra $\sqrt{x}=\dfrac{1}{2} \Leftrightarrow x=\dfrac{1}{4}$
$b)\sqrt{(x-2011)^2+(x-1)^2}\\ =\sqrt{x^2-4022x+4044121+x^2-2x+1}\\ =\sqrt{2x^2 - 4024 x +4044122}\\ =\sqrt{2}\sqrt{x^2 - 2012x +2022061}\\ =\sqrt{2}\sqrt{x^2 - 2.x.1006 +1012036+1010025}\\ =\sqrt{2}\sqrt{(x-1006 )^2+1010025} \ge \sqrt{2}\sqrt{1010025}=1005\sqrt{2} \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow x-1006=0 \Leftrightarrow x=1006$
