-x^2+3x+2
\(=-\left(x^2-3x-2\right)\)
\(=-\left(x^2-3x+\frac{9}{4}-\frac{17}{4}\right)\)
\(=-\left[\left(x-\frac{3}{2}\right)^2-\frac{17}{4}\right]\)
\(=-\left(x-\frac{3}{2}\right)^2+\frac{17}{4}\le\frac{17}{4}\)
=>GTLN khi \(x-\frac{3}{2}=0\)
\(\Rightarrow x=\frac{3}{2}\)
\(-x^2+3x+2=-\left(x^2-3x+\frac{9}{4}\right)+\frac{17}{4}=-\left(x-\frac{3}{2}\right)^2+\frac{17}{4}\le\frac{17}{4}\)
\(x.\left(1-2x\right)=x-2x^2=-2x^2+x=-2\left(x^2-\frac{1}{2}x\right)=-2.\left(x^2-\frac{2.x.1}{4}+\frac{1}{16}\right)+\frac{1}{8}=-2.\left(x-\frac{1}{4}\right)^2+\frac{1}{8}\le\frac{1}{8}\)
tự vt dấu = xảy ra nha :))
