a) \(3-x^2+5x=-\left(x^2-5x-3\right)\)
\(=-\left(x^2-2x.\frac{5}{2}+\frac{10}{4}-\frac{22}{4}\right)\)
\(=-\left(x-\frac{5}{2}\right)^2+\frac{22}{4}\)
\(=-\left(x-\frac{5}{2}\right)^2+\frac{11}{2}\)
Mà: \(\left(x-\frac{5}{2}\right)^2\ge0\)\(\Leftrightarrow-\left(x-\frac{5}{2}\right)^2\le0\)
\(\Leftrightarrow-\left(x-\frac{5}{2}\right)^2+\frac{11}{2}\le\frac{11}{2}\)
\(\Leftrightarrow3-x^2+5x\le\frac{11}{2}\)
Dấu = xảy ra khi: \(\left(x-\frac{5}{2}\right)^2=0\)
\(\Leftrightarrow x-\frac{5}{2}=0\)
\(\Leftrightarrow x=\frac{5}{2}\)(T/m)
Vậy GTLN của 3 - x2 + 5x là \(\frac{11}{2}\)khi x = \(\frac{5}{2}\).
b) \(12-6x^2-6x=-6\left(x^2+x-2\right)\)
\(=-6\left(x^2+2x.\frac{1}{2}+\frac{1}{4}-\frac{9}{4}\right)\)
\(=-6\left(x+\frac{1}{2}\right)^2+\frac{27}{2}\)
Mà: \(\left(x+\frac{1}{2}\right)^2\ge0\)\(\Leftrightarrow-6\left(x+\frac{1}{2}\right)^2\le0\)
\(\Leftrightarrow-6\left(x+\frac{1}{2}\right)^2+\frac{27}{2}\le\frac{27}{2}\)\(\Leftrightarrow12-6x^2-6x\le\frac{27}{2}\)
Dấu = xảy ra khi: \(\left(x+\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow x+\frac{1}{2}=0\)\(\Leftrightarrow x=-\frac{1}{2}\)(T/m)
Vậy GTLN của 12 - 6x2 - 6x là \(\frac{27}{2}\)khi x = \(-\frac{1}{2}\).
