1.ĐKXĐ: \(-2\sqrt{2}\le x\le2\sqrt{2}\)
\(B=\sqrt{16-2x^2}\)
Có: \(x^2\ge0\forall x\Rightarrow2x^2\ge0\forall x\Rightarrow16-2x^2\le16\forall x\)
\(\Rightarrow0\le\sqrt{16-2x^2}\le4\forall-2\sqrt{2}\le x\le2\sqrt{2}\)
Vậy \(B_{max}=4\Leftrightarrow x=0\)
2.
\(A=\sqrt{x^2}+\sqrt{x^2-4x+4}\)
\(A=\sqrt{x^2}+\sqrt{\left(x-2\right)^2}\)
TH1: \(x\ge2\)
\(A=x+x-2=2x-2\)
TH2: \(x< 2\\ A=x+2-x=2\)