Lời giải:
Ta thấy:
\(2(x^2+x+1)=x^2+(x^2+2x+1)+1\)
\(=x^2+(x+1)^2+1\geq 1>0, \forall x\in\mathbb{R}\)
Vậy \(2(x^2+x+1)>0, \forall x\in\mathbb{R}\)
Do đó để \(\frac{2x^2+3x}{2(x^2+x+1)}>0\Leftrightarrow 2x^2+3x>0\)
\(\Leftrightarrow x(2x+3)>0\) \(\left[\begin{matrix} \left\{\begin{matrix} x>0\\ 2x+3>0\end{matrix}\right.\\ \left\{\begin{matrix} x<0\\ 2x+3<0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x>0\\ x< \frac{-3}{2}\end{matrix}\right.\)
Vậy \(x>0\) hoặc \(x< \frac{-3}{2}\)
\(\dfrac{2x^2+3x}{2\left(x^2+x+1\right)}>0\left(1\right)\)
co: \(2\left(x^2+x+1\right)=2\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]=2\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{2}\ge\dfrac{3}{2}\forall x\in R\)
\(\left(1\right)\Leftrightarrow2x^2+3x>0\Leftrightarrow x^2+\dfrac{3}{2}x>0\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right)^2>\dfrac{9}{16}\) \(\Leftrightarrow\left|x+\dfrac{3}{4}\right|>\dfrac{3}{4}\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}< \dfrac{-3}{4}\\x+\dfrac{3}{4}>\dfrac{3}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x< -\dfrac{3}{2}\\x>0\end{matrix}\right.\)