a) \(D=-3x\left(x+3\right)-7=-3x^2-9x-7=-3\left(x^2+3x+\frac{9}{4}\right)-\frac{1}{4}=-3\left(x+\frac{3}{2}\right)^2-\frac{1}{4}\)
Vì: \(-3\left(x+\frac{3}{2}\right)^2\le0\)
=>\(-3\left(x+\frac{3}{2}\right)^2-\frac{1}{4}\le-\frac{1}{4}\)
Vậy GTLN của D là \(-\frac{1}{4}\) khi \(x=-\frac{3}{2}\)
\(D=-3x^2-9x-7=-3\left(x^2+3x+\frac{7}{3}\right)=-3\left[\left(x+\frac{3}{2}\right)^2++\frac{1}{12}\right]\)\(=-3\left(x+\frac{3}{2}\right)^2-\frac{1}{4}\le-\frac{1}{4};\forall x\in R\)
Dấu '=' xảy ra ↔ x + 3/2 = 0 ↔ x = -3/2