Question

b ) \(x^3-3x^2-x+3=0\)

Answer 1

Ta có \(x^3-3x^2-x+3=0\)

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

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

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

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=3\\x=-1\end{array}\right.\)

Answer 2

\(x^3-3x^2-x+3=0\)

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

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

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

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x+1=0\\x-3=0\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\\x=3\end{array}\right.\)

Vậy \(x\in\left\{1;-1;3\right\}\)