Question

\(\frac{1}{3}\le\frac{^{x^2}+x+1}{x^2-x+1}\le3\)CM

Answer 1

\(\frac{1}{3}< =\frac{x^2+x+1}{x^2-x+1}\Rightarrow x^2-x+1< =3x^2+3x+3\Rightarrow x^2-x+1-3x^2-3x-3< =0\)

\(\Rightarrow-2x^2-4x-2< =0\Rightarrow-2\left(x^2+2x+1\right)< =0\Rightarrow-2\left(x+1\right)^2< =0\)

vì \(\left(x+1\right)^2>=0;-2< 0\Rightarrow-2\left(x+1\right)^2< =0\)luôn đúng \(\Rightarrow\frac{1}{3}< =\frac{x^2+x+1}{x^2-x+1}\)luôn dúng (1)

cái kia cx tương tự như vậy nhé