Question

Answer 1

\(\dfrac{2x^3+x^2+1}{x^3+1}=\dfrac{\left(x+1\right)\left(2x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{2x^2-x+1}{x^2-x+1}=\dfrac{16x^2-8x+8}{16x^2-16x+16}\)

\(=\dfrac{\left(4x-1\right)^2+7}{\left(4x-2\right)^2+12}\)

Do \(\left\{{}\begin{matrix}\left(4x-1\right)^2+7\ge7>0\\\left(4x-2\right)^2+12\ge12>0\end{matrix}\right.\)

\(\Rightarrow\dfrac{\left(4x-1\right)^2+7}{\left(4x-2\right)^2+12}>0;\forall x\)

\(\Rightarrow\) Với mọi \(x\ne-1\) thì giá trị của phân thức đã cho luôn dương