Question

C/M: A = x^2 + x + 1 >_ 3/4

Answer 1

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

ta có : \(\left(x+\dfrac{1}{2}\right)^2\ge0\) \(\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(\Leftrightarrow A=x^2+x+1\ge\dfrac{3}{4}\) (đpcm)

Answer 2

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

\(\text{Mà }\left(x+\dfrac{1}{2}\right)^2\ge0\\ \Rightarrow A=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\left(ĐPCM\right)\)

Vậy \(A\ge\dfrac{3}{4}\)