Question

\(\left(2\cdot x^3+3\cdot x^2\right)+\left(3+2\cdot x\right)=0\)

Answer 1

\(\left(2x^3+3x^2\right)+\left(3+2x\right)=0\)

\(\Leftrightarrow x^2\left(2x+3\right)+\left(3+2x\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\x^2+1=0\end{matrix}\right.\)

\(\Leftrightarrow x=-\dfrac{3}{2}\)

Answer 2

\(\left(2.x^3+3.x^2\right)+\left(3+2x\right)=0\)

\(x^2.\left(2x+3\right)+\left(3+2x\right)=0\)

\(\left(3+2x\right).\left(x^2+1\right)=0\)

Ta có: \(x^2+1>0\forall x\)

\(\Rightarrow3+2x=0\)

\(\Rightarrow x=-\dfrac{3}{2}\)

Vậy \(x=-\dfrac{3}{2}\)

Tham khảo nhé~