Question

x² - x +1

tính giá trị nhỏ nhất của x

Answer 1

Đặt

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

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

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

Xét \(\left(x-\dfrac{1}{2}\right)^2\ge0\)

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

Vậy \(Min_A=\dfrac{3}{4}\) . Dấu \(=\) xảy ra khi \(x=\dfrac{1}{2}\)

Answer 2

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

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

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

Dấu "=" xảy ra khi:

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