ko ai trả lời mình tl luôn nha!
\(-2x^2-3x-7=-2\left(x^2+\frac{3}{2}x+\frac{7}{2}\right)\)
\(=-2\left(x^2+2.x.\frac{3}{4}+\frac{9}{16}-\frac{9}{16}+\frac{7}{2}\right)\)
\(=-2\left(x+\frac{3}{4}\right)^2-\frac{47}{8}\le-\frac{47}{8}\)
Dấu "=" xảy ra khi x = -3/4
Vậy..