Question

tìm x:

x³+27+(x+3)(x-9)=0

Answer 1

x³+27+(x+3)(x-9)=0

\(x^3+3^3+\left(x+3\right)\cdot\left(x-9\right)=0\)

\(\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)

\(\left(x+3\right)\cdot\left(x^2-2x\right)=0\)

\(x\cdot\left(x+3\right)\cdot\left(x-2\right)=0\)

\(\left\{{}\begin{matrix}x=0\\x+3=0\\x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)