\(2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow\left(2x^3+2x\right)+\left(3x^2+3\right)=0\)
\(\Leftrightarrow2x\left(x^2+1\right)+3\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\) (Vì \(x^2+1>0\))
\(\Leftrightarrow x=-\frac{3}{2}\)
\(2x^3+3x^2+2x+3=0\)
\(\Rightarrow\left(2x^3+2x\right)+\left(3x^2+3\right)=0\)
\(\Rightarrow\left[2x.\left(x^2+1\right)\right]+\left[3.\left(x^2+1\right)\right]=0\)
\(\Rightarrow\left(x^2+1\right)\left(3+2x\right)=0\)
Suy ra:\(\begin{cases}x^2+1=0\\3+2x=0\end{cases}\)\(\Rightarrow\begin{cases}x\in\varnothing\\x=\frac{-3}{2}\end{cases}\)
Vậy \(x=-\frac{3}{2}\)
\(2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow\left(2x^3+2x\right)+
\left(3x^2+3\right)=0\)
\(\Leftrightarrow2x\left(x^2+1\right)+3\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(2x+3\right)=0\)(1)
\(Có:x^2\ge0\) với mọi x
\(\Rightarrow x^2+1\ge0+1=1\) với mọi x
\(\Rightarrow x^2+1>0\) với mọi x
(1)\(\Leftrightarrow2x+3=0\)
\(\Leftrightarrow x=\frac{-3}{2}\)