x3 + 3x - 4 = 0=> x3 - x2 + x2 - x + 4x - 4 = 0=> x2 ( x - 1 ) + x ( x - 1 ) + 4 ( x - 1 ) = 0=> ( x2 + x + 4 ) ( x - 1 ) = 0\(\Leftrightarrow\orbr{\begin{cases}x^2+x+4=0\\x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x^2+x=-4\left(loai\right)\\x=1\end{cases}}\)Câu trả lời: x3 + 3x - 4 = 0=> x3 - x2 + x2 - x + 4x - 4 = 0=> x2 ( x - 1 ) + x ( x - 1 ) + 4 ( x - 1 ) = 0=> ( x2 + x + 4 ) ( x - 1 ) = 0\(\Leftrightarrow\orbr{\begin{cases}x^2+x+4=0\\x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x^2+x=-4\left(loai\right)\\x=1\end{cases}}\)