Với dạng bài này ta chỉ việc chia hoocne là ra nhé!

\(C1:x^4+x^3-8x^2-9x-9=0\\ \Leftrightarrow\left(x-3\right)\left(x^3+4x^2+4x+3\right)\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x^2+x+1\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=0\\x^2+x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x^2+x+1=0\left(VN\right)\end{matrix}\right.\)

\(C2:x^4+2x^3-3x^2-8x-4=0\\ \Leftrightarrow\left(x+1\right)\left(x^3+x^2-4x-4\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x+1\right)\left(x^2-4\right)=0\\ \Leftrightarrow\left(x+1\right)^2\left(x^2-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)^2=0\\x^2-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-2\end{matrix}\right.\)

ta có : (x^2+x+4)(1+8x) +16x^2=0

vì 16x^2>=0 suy ra *x^2+x+4=0

*1+8x=0

*16x^2=0

tự giải pt

Lời giải:

\((x^3-x^2)-4x^2+8x-4=0\)

\(\Leftrightarrow x^2(x-1)-4(x^2-2x+1)=0\)

\(\Leftrightarrow x^2(x-1)-4(x-1)^2=0\)

\(\Leftrightarrow (x-1)[x^2-4(x-1)]=0\)

\(\Leftrightarrow (x-1)(x^2-4x+4)=0\)

\(\Leftrightarrow (x-1)(x-2)^2=0\)

\(\Rightarrow \left[\begin{matrix} x=1\\ x=2\end{matrix}\right.\)