1)
\(A=x^2-x+3\\=\bigg[x^2-2\cdot x\cdot \dfrac{1}{2} +\bigg(\dfrac{1}{2}\bigg)^2\bigg]-\dfrac{1}{4} +3\\=\bigg(x-\dfrac{1}{2}\bigg)^2+\dfrac{11}{4}\)
Ta thấy: \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{11}{4}\le\dfrac{11}{4}\forall x\)
Dấu \("="\) xảy ra khi: \(x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(Min_A=\dfrac{11}{4}\) khi \(x=\dfrac{1}{2}\).
2)
\(B=x^2+x+1\\=\bigg[x^2+2\cdot x\cdot \dfrac{1}{2}+\bigg(\dfrac{1}{2}\bigg)^2\bigg]-\dfrac{1}{4}+1\\=\bigg(x+\dfrac{1}{2}\bigg)^2+\dfrac{3}{4}\)
Ta thấy: \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\le\dfrac{3}{4}\forall x\)
Dấu \("="\) xảy ra khi: \(x+\dfrac{1}{2}=0\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy \(Min_B=\dfrac{3}{4}\) khi \(x=-\dfrac{1}{2}\).
\(Toru\)
