求解几道初中代数题,最好附上解释,急!n(n+1)(n+2)-3n(n+1) =(x+1)(x-2)+(2-x)(3-x) =a2+4ab+4b2-c2 =4-4m2+12mn-9n2=9+x2-z2-6x请用因式分解

1.原式=n(n+1)[(n+2)-3]=n(n+1)(n-1)
2.原式=(x-2)[(x+1)+(x-3)]=2(x-2)(x-1)
3.原式=(a+2b)^2-c^2=(a+2b+c)(a+2b-c)
4.原式=4-(4m^2-12mn+9n^2)=2^2-(2m-3n)^2=(2+2m-3n)(2-2m+3n)
5.原式=x^2-6x+9-z^2=(x-3)^2-z^2=(x-3-z)(x-3+z)

第一个=（n+1)(n*n+2*n-3*n)=n(n+1)(n-1)
第2个=（x-2)(x+1+x-3)=2(x-2)(x-1)
.....=(a+2b)^2-c^2=(a+2b+c)(a+2b-c)
....=4-(2m-3n)^2=(2+2m-3n)(2-2m+3n)
.....=9+x^2-6x-z^=(x-3)^2-z^2=(x-3+z)(x-3-z)
全手工 给分

1、原式=(n+1)[n(n+2)-3n]=(n+1)(n^2-n)=n(n-1)(n+1)
2、原式=(x-2)[x+1-(3-x)]=(x-2)(2x-2)=2(x-2)(x-1)
3、原式=(a+2b)^2-c^2=(a+2b+c)(a+2b-c)
4、原式=4-(2m-3n)^2=(2+2m-3n)(2-2m+3n)
5、原式=(x-3)^2-z^2=(x-3+z)(x-3-z)