a:Ta có: \(A=-4x^2+x-1\)
\(=-4\left(x^2-\dfrac{1}{4}x+\dfrac{1}{4}\right)\)
\(=-4\left(x^2-2\cdot x\cdot\dfrac{1}{8}+\dfrac{1}{64}+\dfrac{63}{64}\right)\)
\(=-4\left(x-\dfrac{1}{8}\right)^2-\dfrac{63}{16}\le-\dfrac{63}{16}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{8}\)
b: Ta có: \(B=-3x^2+5x+6\)
\(=-3\left(x^2-\dfrac{5}{3}x-2\right)\)
\(=-3\left(x^2-2\cdot x\cdot\dfrac{5}{6}+\dfrac{25}{36}-\dfrac{97}{36}\right)\)
\(=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{97}{12}\le\dfrac{97}{12}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{5}{6}\)
c: Ta có: \(C=-x^2+3x+4\)
\(=-\left(x^2-3x-4\right)\)
\(=-\left(x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{25}{4}\right)\)
\(=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)
