\(\Leftrightarrow x^3\left(x^2-1\right)-\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^3-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow x=\pm1\)
x5 - x3 - x2 + 1 = 0
\(\Leftrightarrow\) (x5 - x2) - (x3 - 1) = 0
\(\Leftrightarrow\) x2(x3 - 1) - (x3 - 1) = 0
\(\Leftrightarrow\) (x3 - 1)(x2 - 1) = 0
\(\Leftrightarrow\) (x-1)(x2 + x +1) (x-1)(x+1) = 0
\(\Leftrightarrow\) (x-1)2 \(\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]\) (x+1) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]\\x+1=0\end{matrix}\right.=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left(x+\dfrac{1}{2}\right)^2\\x=-1\end{matrix}\right.=\dfrac{-3}{4}\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\varnothing\\x=-1\end{matrix}\right.\)
Vậy x=1;x= -1 là nghiệm của pt
f(1) =0> x=1 là nghiệm
f(-1) =0> x=-1 là nghiệm
chia đa thức được (x^2 -1)
<=>(x^2-1) (x^3 -1) =0
x^3 -1 =(x-1) (x^2 +x+1)
có x^2 +x+1 >0
=> x =+-1
