b) Ta có: \(x^3-x^2+x-1=0\)
\(\Leftrightarrow x^2\left(x-1\right)+\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+1\right)=0\)(1)
Ta có: \(x^2\ge0\forall x\)
\(\Rightarrow x^2+1\ge1\ne0\forall x\)(2)
Từ (1) và (2) suy ra \(x-1=0\)
\(\Leftrightarrow x=1\)
a) \(x^3-x^2-4=0\Leftrightarrow\left(x-2\right)\left(x^2+x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x^2+x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x\in\varnothing\end{matrix}\right.\)
Vậy $x=2$
b) \(x^3-x^2+x-1=0\Leftrightarrow\left(x-1\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x\in\varnothing\end{matrix}\right.\)
Vậy $x=1$
c) \(x^3+x^2+4=0\Leftrightarrow\left(x+2\right)\left(x^2-x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x^2-x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\end{matrix}\right.\)
Vậy $x=-2$
