a)Đặt \(A=\sqrt{x-2}+\sqrt{4-x}\)
Đk:\(2\le x\le4\)
\(A^2=x-2+4-x+2\sqrt{\left(x-2\right)\left(4-x\right)}\)
\(=2+2\sqrt{\left(x-2\right)\left(4-x\right)}\) (dùng BĐT Cauchy)
\(\le2+\left(x-2\right)+\left(4-x\right)\)
\(=2+2=4\)
\(\Rightarrow A^2\le4\Leftrightarrow A\le2\)
Dấu = khi \(\sqrt{x-2}=\sqrt{4-x}\Leftrightarrow x=3\)
Vậy MaxA=2 khi x=3
b)Đặt \(B=\sqrt{6-x}+\sqrt{x+2}\)
Đk:\(-2\le x\le6\)
\(B^2=6-x+x+2+2\sqrt{\left(6-x\right)\left(x+2\right)}\)
\(=8+2\sqrt{\left(6-x\right)\left(x+2\right)}\) (Bđt Cauchy)
\(\le8+\left(6-x\right)+\left(x+2\right)\)
\(=8+8=16\)
\(\Rightarrow B^2\le16\Leftrightarrow B\le4\)
Dấu = khi \(\sqrt{6-x}=\sqrt{x+2}\Leftrightarrow x=2\)
Vậy MaxB=4 khi x=2
c)Đặt \(C=\sqrt{x}+\sqrt{2-x}\)
Đk:\(0\le x\le2\)
\(C^2=x+2-x+2\sqrt{x\left(2-x\right)}\)
\(=2+2\sqrt{x\left(2-x\right)}\) (bđt Cauchy)
\(\le2+x+\left(2-x\right)\)
\(=2+2=4\)
\(\Rightarrow C^2\le4\Leftrightarrow C\le2\)
Dấu = khi \(\sqrt{x}=\sqrt{2-x}\Leftrightarrow x=1\)
Vậy MaxC=2 khi x=1
