Question

Tìm min

B= x^2 - x +1

Answer 1

\(B=x^2-x+1\)

\(B=x^2-2\cdot\dfrac{1}{2}\cdot x+\dfrac{1}{4}+\dfrac{3}{4}\)

\(B=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

Mà: \(\left(x-\dfrac{1}{2}\right)^2\ge0\) nên \(B=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Dấu "=" xảy ra:

\(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=\dfrac{3}{4}\)

\(\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)

Vậy: \(B_{min}=\dfrac{3}{4}\)khi \(x=\dfrac{1}{2}\)

Answer 2

B=x^2-x+1

=x^2-x+1/4+3/4

=(x-1/2)^2+3/4>=3/4

Dấu = xảy ra khi x=1/2