a) \(A=\sqrt{x-3}-3\ge-3\)
\(minA=-3\Leftrightarrow x=3\)
b) \(B=x-2\sqrt{x}+5=\left(x-2\sqrt{x}+1\right)+4=\left(\sqrt{x}-1\right)^2+4\ge4\)
\(minB=4\Leftrightarrow x=1\)
c) \(C=x-3\sqrt{x}=\left(x-3\sqrt{x}+\dfrac{9}{4}\right)-\dfrac{9}{4}=\left(\sqrt{x}-\dfrac{3}{2}\right)^2-\dfrac{9}{4}\ge-\dfrac{9}{4}\)
\(minC=-\dfrac{9}{4}\Leftrightarrow x=\dfrac{9}{4}\)
d) \(D=\sqrt{x^2-4x+5}+2=\sqrt{\left(x^2-4x+4\right)+1}+2=\sqrt{\left(x-2\right)^2+1}+2\ge\sqrt{1}+2=3\)
\(minD=3\Leftrightarrow x=2\)
Bài 2:
\(P=\dfrac{\sqrt{x}-2}{\sqrt{x}+5}=1-\dfrac{7}{\sqrt{x}+5}\le-\dfrac{2}{5}\forall x\)
Dấu '=' xảy ra khi x=0
