Question

Giải các phương trình sau:

a) \(x^3-3x^2-4x=0\)

b) \(3x^2-5x-2=0\)

Answer 1

a) \(x^3-3x^2-4x=0\)

\(\Leftrightarrow x\left(x^2-3x-4\right)=0\)

\(\Leftrightarrow x\left(x-4\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-1\end{matrix}\right.\)

Vậy \(S=\left\{0;4;-1\right\}\).

b) \(3x^2-5x-2=0\)

\(\Leftrightarrow3x^2+x-6x-2=0\)

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

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

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

Vậy \(S=\left\{-\dfrac{1}{3};2\right\}\).