$a)_{}$ $x.(3x+2)+(x+1)^2-(2x-5)(2x+5)=-12_{}$
$⇔3x^2+2x+x^2+2x+1-(4x^2-25)=-12_{}$
$⇔3x^2+2x+x^2+2x+1-4x^2+25=-12_{}$
$⇔4x+26=-12_{}$
$⇔4x=-38_{}$
$⇔x_{}=$ $-\frac{19}{2}$
$Vậy_{}$ $x=_{}$ $-\frac{19}{2}$
$b)_{}$ $x^3+27+(x+3)(x-9)=0_{}$
$⇔x^3+27+x^2-9x+3x-27=0_{}$
$⇔x^3+x^2-6x=0_{}$
$⇔x.(x^2+x-6)=0_{}$
$⇔x.(x^2+3x-2x-6)=0_{}$
$⇔x.[ x.(x+3)-2.(x+3)]=0 _{}$
$⇔x.(x+3)(x-2)=0_{}$
Đến đây tìm x được rồi nhé
$Vậy_{}$ $x_{1}=0;$ $x_{2}=-3$ $;x_{3}=2$
