Question

P=\(\sqrt{x^2-x+1}\)

tìm GTLN của P

Answer 1

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 Min_P = \(\frac{\sqrt{3}}{2}\) khi \(x-\frac{1}{2}=0\) <=> x = 1/2 .

Answer 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 = \(\frac{1}{2}\)