Question

I : Tìm min

F=\(\sqrt{x^2+2019}\)

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

help me !!!

Answer 1

F=\(\sqrt{x^2+2019}\)

=>\(F^2=x^2+2019 =>x^2+2019\)≥2019

=> \(F^2 \)_min=2019=>F _min=\(\sqrt{2019}\)<=>x=0

Answer 2

G=\(\sqrt{x^2-x+1}\)=\(\sqrt{x^2-2.\frac{1}{2}.x+\frac{1}{4}+\frac{3}{4}}\)=\(\sqrt{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}\) \(\ge\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}\)

Dấu "=" xảy ra <=> x=\(\frac{1}{2}\)

Vậy minG=\(\frac{\sqrt{3}}{2}\) <=> x\(=\frac{1}{2}\)