设a(x^3y^m)^2÷(3x^ny^4)^2=2x^5y^4,则a=?m=?n=?

a(x^3y^m)^2/(3x^ny^4)^2
=ax^6y^2m/(9x^2ny^8)
=(a/9)*x^(6-2n)*y^(2m-8)
=2x^5y^4
所以a/9=2
6-2n=5
2m-8=4

a=18,m=6,n=1/2

我靠
看见这东西都头疼

a(x^3y^m)^2÷(3x^ny^4)^2
=(a/9)*x^(3*2-2n)*y^(2m-4*2)
=(a/9)*x^(6-2n)*y^(2m-8)

所以：a/9=2，6-2n=5，2m-8=4
解得：a=18，n=1/2，m=6

因为a(x^3y^m)^2÷(3x^ny^4)^2=2x^5y^4,
所以ax^6y^(2m)/9x^(2n)y^8=2x^5y^4,
所以(a/9)x^(6-2n)y^(2m-8)=2x^5y^4,
所以比较系数和指数得a/9=2,6-2n=5,2m-8=4,
所以a=18,n=1/2,m=6.

a=18,
m=6
n=1/2

多练习下吧