Áp dụng BĐT Bunhiacopxki ta có :
\(C=\sqrt{x}+\sqrt{2-x}\le\sqrt{\left(1+1\right)\left(x+2-x\right)}\)
\(\Leftrightarrow C\le\sqrt{2.2}\)
\(\Leftrightarrow C\le2\)
Dấu "=" xảy ra \(\Leftrightarrow x=2-x\)
\(\Leftrightarrow x=1\)
Vậy...
ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\2-x\ge0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge0\\x\le2\end{matrix}\right.\)
=> \(0\le x\le2\)
Ta có : \(C=1\sqrt{x}+1\sqrt{2-x}\)
Ta thấy : \(1\sqrt{x}+1\sqrt{2-x}\le\sqrt{\left(1^2+1^2\right)+\left(\left(\sqrt{x}\right)^2+\left(\sqrt{2-x}\right)^2\right)}\)
=> \(C\le\sqrt{2+x+2-x}=\sqrt{4}=2\)
Vậy MaxC = 2 <=> \(\frac{1}{\sqrt{x}}=\frac{1}{\sqrt{2-x}}\)
<=> \(x=2-x\)
<=> \(x=1\) ( TM )