Ta có :
\(F=3x-2x^2\)
\(\Leftrightarrow F=-2\left(x^2-\frac{3}{2}x+\frac{9}{16}\right)+\frac{9}{8}\)
\(\Leftrightarrow F=\frac{9}{8}-2\left(x-\frac{3}{4}\right)^2\)
Với mọi x ta có :
\(\left(x-\frac{3}{4}\right)^2\ge0\)
\(\Leftrightarrow-2\left(x-\frac{3}{4}\right)^2\le0\)
\(\Leftrightarrow\frac{9}{8}-2\left(x-\frac{3}{4}\right)^2\le\frac{9}{8}\)
\(\Leftrightarrow F\le\frac{9}{8}\)
Dấu "=" xảy ra \(\Leftrightarrow x=\frac{3}{4}\)
Vậy....................
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