ĐK: \(-1\le x\le\frac{1}{2}\)
\(B=\frac{x}{2}+\sqrt{1-x-2x^2}\)
\(B=\frac{x}{2}+\sqrt{\left(x+1\right)\left(1-2x\right)}\)
Áp dụng bđt Cô-si :
\(B\le\frac{x}{2}+\frac{x+1+1-2x}{2}=\frac{x}{2}+\frac{2-x}{2}=\frac{x+2-x}{2}=1\)
Dấu "=" xảy ra \(\Leftrightarrow x+1=1-2x\Leftrightarrow x=0\)