Giải thích các bước giải:
a.Ta có:
$x-2\sqrt{x}-2=x-2\sqrt{x}+1-3=(\sqrt{x}-1)^2-3\ge -3$
Dấu = xảy ra khi $x=1$
b.Ta có:
$x-6\sqrt{x}+1=x-6\sqrt{x}+9-8=(\sqrt{x}-3)^2-8\ge -8$
Dấu = xảy ra khi $\sqrt{x}-3=0\to x=9$
c.Ta có:
$x-\sqrt{x}+2=x-\sqrt{x}+\dfrac14+\dfrac74=(\sqrt{x}-\dfrac12)^2+\dfrac74\ge \dfrac74$
Dấu = xảy ra khi $\sqrt{x}-\dfrac12=0\to x=\dfrac14$
d.Ta có:
$2x-3\sqrt{x}+2=2(x-2\sqrt{x}\cdot\dfrac34+\dfrac9{16})+\dfrac78=2(\sqrt{x}-\dfrac34)^2+\dfrac78\ge \dfrac78$
Dấu = xảy ra khi $\sqrt{x}-\dfrac34=0\to x=\dfrac9{16}$
e.Ta có:
$x+\dfrac{4}{\sqrt{x}}+1$
$=x+\dfrac{2}{\sqrt{x}}+\dfrac{2}{\sqrt{x}}+1$
$\ge 3\sqrt[3]{x.\dfrac{2}{\sqrt{x}}.\dfrac{2}{\sqrt{x}}}+1$
$\ge 3\sqrt[3]{4}+1$
Dấu = xảy ra khi $x=\dfrac{2}{\sqrt{x}}=\sqrt[3]{4}$
