Ta có : a , \(-x^2+3x-3=-\left(x^2-3x+3\right)=-\left(x^2-2.x.\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{3}{4}\right)=-\left[\left(x-\dfrac{3}{2}\right)^2+\dfrac{3}{4}\right]< 0,\forall x\)
b , \(-3x^2+6x+4\) = \(-\left(3x^2-6x-4\right)=-3\left(x^2-2x-\dfrac{4}{3}\right)=-3\left(x^2-2x+1-\dfrac{7}{3}\right)=-3\left[\left(x-1\right)^2-\dfrac{7}{3}\right]\)
============== câu b á ... ko chắc nó < 0 được
Bạn thử x=1 thì biết ah
a)
\(-\left(x^2-2.\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{3}{4}\)
\(=-\left(x-\dfrac{3}{2}\right)^2-\dfrac{3}{4}\le\dfrac{-3}{4}< 0\)
