三道初一计算题（过程）1.(2m^2-n)^2(2m^2+n)^22.(x-3y)(x+3y)-(x-3y)^23.若(x^2+nx+3)(x^2-3x+m)的乘积中不含x^2和x^3项,求m和n的值.

1.原式=[(2m^2-n)(2m^2+n)]^2=(4m^4-n^2)^2=16m^8-8m^4n^2+n^4
2.原式=x^2-9y^2-x^2+6xy-9y^2=6xy-18y^2
3.原式=x^4+(n-3)x^3+(m-3n+3)x^2+(mn-9)x+3m，因为不含x^2和x^3，故n-3=0，m-3n+3=0，解得：m=6，n=3

1）
=[(2m^2-n)(2m^2+n)]^2
=[(2m^2)^2-n^2]
=4m^4-n^2
2)
=(x-3y)*[(x+3y)-(x-3y)]
=(x-3y)*6y
=6xy-18y^2
3)(x^2+nx+3)(x^2-3x+m)中
x^2的系数[1*m+n*(-3)+3*1]=0
x^3的系数[1*(-3)+n*1]=0
得m=6 n=3

1.(2m^2-n)^2(2m^2+n)^2
=[(2m^2-n)(2m^2+n)]^2
=(4m^4-n^2)^2
=16m^8-8m^4n^2+n^4
2.(x-3y)(x+3y)-(x-3y)^2
=(x-3y)[(x+3y)-(x-3y)]
=(x-3y)6y
=6xy-18y^2
3.(x^2+nx+3)(x^2-3x+m)
=x^4-(3-n)x^3+(m-3n+3)x^2+mnx+3m
当m=6
n=3
就能把x^2和x^3项消去了
应该能看懂吧