\(A=x-2\sqrt{x+2}=x+2-2\sqrt{x+2}+1-3=\left(\sqrt{x+2}-1\right)^2-3\ge-3\)
Dấu "=" xảy ra khi \(\sqrt{x+2}-1=0\Leftrightarrow\sqrt{x+2}=1\Leftrightarrow x+2=1\Leftrightarrow x=-1\)
Vậy \(Min_A=-3\) khi x=-1
A=x−2√(x+2) (ĐK:x≥-2)
A=(x+2)-2√(x+2)+1-3
A=\(\left(\sqrt{x+2}-1\right)^2-3\)
Ta có (\(\left(\sqrt{x+2}-1\right)^2\)≥0
<=>\(\left(\sqrt{x+2}-1\right)^2-3\)≥-3
<=> A≥3
Dấu = xảy ra<=>\(\left(\sqrt{x+2}-1\right)^2\)=0
<=> \(\sqrt{x+2}\)-1=0
<=>\(\sqrt{x+2}\)=1
<=> x+2=1
<=> x=-1 (tm)
Vậy GTNN A=-3 khi x=-1
