b)
\(-x^2+3x-2=-\left(x+\dfrac{3}{-2}\right)^2+\dfrac{3.\left(-1\right).\left(-2\right)-9}{2.\left(-2\right)}\\ =-\left(x+\dfrac{3}{-2}\right)^2+\dfrac{1}{4}\)
vì \(-\left(x+\dfrac{3}{-2}\right)^2\le0\) nên
\(-\left(x+\dfrac{3}{-2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
vậy MAXA = 0,25 tại x=1,5