\(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)
\(x^3-3^3+x\left(2^2-x^2\right)=1\)
\(x^3-27+4x-x^3=1\)
\(4x-27=1\)
\(4x=28\)
\(x=7\)
Vậy x = 7
\(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)
\(\Rightarrow x^3-3^3+x\left(2^2-x^2\right)=1\)
\(\Rightarrow x^3-27+4x-x^3=1\)
\(\Rightarrow4x-27=1\)
\(\Rightarrow4x=28\)
\(\Rightarrow x=7\)
Vậy \(x=7\)