Question

Tìm GTLN của B = \(1-x^2+3x\)

Answer 1

B = $1-x^2+3x$

= -(x²-3x-1)

= -\(\left(x^2-3x+\dfrac{9}{4}-\dfrac{5}{4}\right)\)

= -\(\left[\left(x-\dfrac{3}{2}\right)^2-\dfrac{5}{4}\right]\)

= \(-\left(x-\dfrac{3}{2}\right)^2+\dfrac{5}{4}\)

do \(-\left(x-\dfrac{3}{2}\right)^2\le0\forall x\)

⇔\(-\left(x-\dfrac{3}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\)

⇔B ≤ \(\dfrac{5}{4}\)

Max A =\(\dfrac{5}{4}\) dấu "=" xảy ra khi \(x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{2}\)

vậy GTLN A = \(\dfrac{5}{4}\) khi x= \(\dfrac{3}{2}\)