Question

tìm GTNN

c) D = x²-x+\(\dfrac{1}{2}\)

Answer 1

d: Ta có: \(D=x^2-x+\dfrac{1}{2}\)

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

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

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

Answer 2

D= (x^2 -2x1/2 +1/4) +1/4

   =(x-1/2)^2 +1/4

MinD=1/4 khi x=1/2

Answer 3