\(x^4+2x^3-2x^2+2x-3=0\Leftrightarrow\left(x-1\right)\left(x+3\right)\left(x^2+1\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}\)
x^4-x^3+3x^3-3x^2+x^2-x+3x-3=0
(x^3+3x^2+x+1)(x-1)=0
.......
x4 + 2x3 - 2x2 - 3 =0
(=) x4 + 3x3 - x3 - 3x2 + x2 + 3x -x - 3 =0
(=) x3(x+3) - x2(x+3) + x(x+3) - (x+3) = 0
(=) (x+3) . (x3 - x2 + x -1 ) =0
(=) (x+3) . [x2(x-1) + (x-1)] = 0
(=) (x+3).(x-1).(x2 + 1) = 0
TH1 : x2 + 1 = 0 TH2: \(\orbr{\begin{cases}x+3=0\\x-1=0\end{cases}}\)(=) \(\orbr{\begin{cases}x=-3\\x=1\end{cases}}\)
(=) x2 = -1 (loại )
Vậy phương trình có nghiệm là : x = -3 , x = 1
