a) \(3x\left(x^2+2x-5\right)-\left(x+4\right)\left(3x^2-1\right)+6x^2=18\)
\(\Rightarrow3x^3+6x^2-15x-3x^3-12x^2+x+4+6x^2=18\)
\(\Rightarrow-14x=14\)
\(\Rightarrow x=-1\)
b, \(\left(x+3\right)^2+\left(x+2\right)\left(x^2-2x+4\right)=x^2+17\)
\(\Rightarrow x^2+6x+9+x^3+8=x^2+17\)
\(\Rightarrow x^3+6x=0\)
\(\Rightarrow x\left(x^2+6\right)=0\)
Vì \(x^2+6>0\Rightarrow x=0\)