\(f,\Leftrightarrow x^3+2x^2+5x-2x^2-4x-10+2\left(x^2-4\right)-5x+10=0\\ \Leftrightarrow x^3-4x+2x^2-8=0\\ \Leftrightarrow x^3+2x^2-4x-8=0\\ \Leftrightarrow x^2\left(x-2\right)-4\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
a) \(3x\left(x-2\right)-x+2=0\)
\(3x\left(x-2\right)-\left(x-2\right)=0\)
\(\left(x-2\right)\left(3x-1\right)=0\)
⇔\(\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
a) \(3x\left(x-2\right)-x+2=0\)
\(\Leftrightarrow3x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=2\end{matrix}\right.\)
b)\(x^2\left(x+1\right)+2x\left(x+1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)
a: Ta có: \(3x\left(x-2\right)-x+2=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
b: Ta có: \(x^2\left(x+1\right)+2x\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\cdot x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\\x=0\end{matrix}\right.\)