Question

tìm GTLN B=\(^{-2x^2-3x+5}\)

 

Answer 1

\(B=-2\left(x^2+\dfrac{3}{2}x+\dfrac{9}{16}\right)+\dfrac{49}{8}=-2\left(x+\dfrac{3}{4}\right)^2+\dfrac{49}{8}\le\dfrac{49}{8}\)

\(B_{max}=\dfrac{49}{8}\) khi \(x=-\dfrac{3}{4}\)

Answer 2

\(B=-2x^2-3x+5=-2\left(x^2+\dfrac{3}{2}x+\dfrac{9}{16}\right)+\dfrac{49}{8}=-2\left(x+\dfrac{3}{4}\right)^2+\dfrac{49}{8}\le\dfrac{49}{8}\)

\(maxB=\dfrac{49}{8}\Leftrightarrow x=-\dfrac{3}{4}\)

Answer 3

\(B=-2x^2-3x+5\)

\(=-2\left(x^2+\dfrac{3}{2}x-\dfrac{5}{2}\right)\)

\(=-2\left(x^2+2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{49}{16}\right)\)

\(=-2\left(x+\dfrac{3}{4}\right)^2+\dfrac{49}{8}\ge\dfrac{49}{8}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{3}{4}\)