\(3x^2-15x^2+2x-10< 0\)
\(\Leftrightarrow3x^2\left(x-5\right)+2\left(x-5\right)< 0\)
\(\Leftrightarrow\left(3x^2+2\right)\left(x-5\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}x-5< 0\\3x^2+2< 0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x< 5\\3x^2< -2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x< 5\\x^2< \dfrac{-2}{3}\end{matrix}\right.\)Vì \(x^2\ge\dfrac{-2}{3}\)
Vậy x = 5
\(3x^3-15x^2+2x-10< 0\)
\(\Leftrightarrow3x^2\left(x-5\right)+2\left(x-5\right)< 0\)
\(\Leftrightarrow\left(3x^2+2\right)\left(x-5\right)< 0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x^2+2< 0\\x-5>0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3x^2+2>0\\x-5< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\in\varnothing\\x>5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x\in R\\x< 5\end{matrix}\right.\)
