Question

tìm GTLN

a,x-x²

b,4x-x²+3

Answer 1

a)\(x-x^2=-x^2+x-\dfrac{1}{4}+\dfrac{1}{4}\)

\(=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)

Đẳng thức xảy ra khi \(x=\dfrac{1}{2}\)

b)\(4x-x^2+3=-x^2+4x-4+7\)

\(=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)

Đẳng thức xảy ra khi \(x=2\)