b/ $(c+3)(c-2)(c+1)\\=[(c+3)(c-2)](c+1)\\=[c(c-2)+3(c-2)](c+1)\\=[c^2-2c+3c-6](c+1)\\=(c^2+c-6)(c+1)\\=(c^2+c-6)c+(c^2+c-6).1\\=c^3+c^2-6c+c^2+c-6\\=c^3+(c^2+c^2)+(-6c+c)-6\\=c^3+2c^2-5c-6$
Vậy $(c+3)(c-2)(c+1)=c^3+2c^2-5c-6$
a/ $(x+3)(x-1)-x(x-5)=11\\↔x(x-1)+3(x-1)-x^2+5x=11\\↔x^2-x+3x-3-x^2+5x=11\\↔(x^2-x^2)+(-x+3x+5x)-3=11\\↔7x=14\\↔x=2$
Vậy $x=2$
