Ta có : \(x^4+2x^3-3x^2-8x-4=0\)
=> \(x^4-2x^3+4x^3-8x^2+5x^2-10x+2x-4=0\)
=> \(x^3\left(x-2\right)+4x^2\left(x-2\right)+5x\left(x-2\right)+2\left(x-2\right)=0\)
=> \(\left(x^3+4x^2+5x+2\right)\left(x-2\right)=0\)
=> \(\left(x^3+x^2+3x^2+3x+2x+2\right)\left(x-2\right)=0\)
=> \(\left(x^2\left(x+1\right)+3x\left(x+1\right)+2\left(x+1\right)\right)\left(x-2\right)=0\)
=> \(\left(x^2+3x+2\right)\left(x+1\right)\left(x-2\right)=0\)
=> \(\left(x^2+x+2x+2\right)\left(x+1\right)\left(x-2\right)=0\)
=> \(\left(x\left(x+1\right)+2\left(x+1\right)\right)\left(x+1\right)\left(x-2\right)=0\)
=> \(\left(x+1\right)\left(x+2\right)\left(x+1\right)\left(x-2\right)=0\)
=> \(\left(x+1\right)^2\left(x+2\right)\left(x-2\right)=0\)
=> \(\left[{}\begin{matrix}\left(x+1\right)^2=0\\x+2=0\\x-2=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-1\\x=-2\\x=2\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{-1,-2,2\right\}\)
x4 + 2x3 - 3x2 -8x - 4 = 0
⇔ x4 + 2x3 - 2x2 - 4x - x2 - 4x - 4 = 0
⇔ x3(x + 2) - 2x(x + 2) - (x + 2)2 = 0
⇔ (x + 2)(x3 - 2x - 1) = 0
⇔ (x - 2)(x3 - x - x - 1) =
⇔ (x - 2)[x(x2 - 1) - (x + 1)] = 0
⇔ (x - 2)(x + 1)(x2 + x - 1) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\text{⇔}\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\) (Vì x2 + x - 1 > 0)
Vậy phương trình có tập nghiệm S={2;-1}
