Question

Phân tích đa thức thành nhân tử 4x^8-4x^2y^6

Tìm x (x^3-4x^2) - x+4=0

Help me !!!

Answer 1

1) \(4x^8-4x^2y^6\)

\(=4x^2\left(x^6-y^6\right)\)

\(=4x^2\left[\left(x^2\right)^3-\left(y^2\right)^3\right]\)

\(=4x^2\left(x^2-y^2\right)\left[\left(x^2\right)^2+x^2y^2+\left(y^2\right)^2\right]\)

\(=4x^2\left(x-y\right)\left(x+y\right)\left(x^4+x^2y^2+y^4\right)\)

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

\(\Rightarrow x^2\left(x-4\right)-\left(x-4\right)=0\)

\(\Rightarrow\left(x-4\right)\left(x^2-1\right)=0\)

\(\Rightarrow\left(x-4\right)\left(x-1\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=1\\x=-1\end{matrix}\right.\)