\(x^3-x^2-x+1=0\)
\(\Leftrightarrow\left(x^3-x^2\right)-\left(x-1\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\mp1\end{matrix}\right.\)
\(\Rightarrow S=\left\{-1;1\right\}\)
\(x^3-x^2-x+1=0\)
\(\Leftrightarrow x^3+x^2-2x^2-2x+x+1=0\)
\(\Leftrightarrow x^2\left(x+1\right)-2x\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\)
suy ra x = -1 hoặc x = 1
x3 -x2-x+1 = 0
⇔(x3-x2)-(x-1)=0
⇔x2(x-1)-(x-1)=0
⇔(x-1)(x2-1)=0
⇔(x-1)(x-1)(x+1)=0
\(\)\(\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
(x3-x2)-(x-1)=0
=> x2(x-1)-(x-1)=0
=> (x2-1)(x-1)=0
=> (x-1)(x+1)(x-1)=0
=> (x-1)2(x+1)=0
=> \(\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
X^3 - x^2 -x+ 1 = 0
( x^3 - x^2 ) - ( x-1) = 0
X^2 ×( x-1) - (x -1) =0
(X-1) × ( x^2 -1) =0
(X-1) ×(x-1) ×( x+1) = 0
X-1 =0
X+1=0
X -1= 0
X = { 1; -1}
Tập nghiệm của phương trình S={ 1; -1}
