Ta có : \(P=\sqrt{x^2-x+1}\)
=> \(P=\sqrt{x^2-2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}}\)
=> \(P=\sqrt{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}\)
- Ta thấy : \(\left(x-\frac{1}{2}\right)^2\ge0\)
=> \(\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
=> \(\sqrt{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}\ge\sqrt{\frac{3}{4}}\)
Vậy MinP = \(\frac{\sqrt{3}}{2}\) khi \(x-\frac{1}{2}=0\) <=> x = 1/2 .
Ta có P = \(\sqrt{x^2-x+1}\) = \(\sqrt{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}\) (P luôn xác định)
Xét P = \(\sqrt{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}\) ≥ \(\sqrt{\frac{3}{4}}\)
Nên Min P = \(\sqrt{\frac{3}{4}}\) tại x = \(\frac{1}{2}\)