Question

Giúp e dạng này vs

Answer 1

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) \(D=\sqrt{x^2-4x+5}+2=\sqrt{\left(x^2-4x+4\right)+1}+2=\sqrt{\left(x-2\right)^2+1}+2\ge\sqrt{1}+2=3\)

\(minD=3\Leftrightarrow x=2\)

Answer 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