x=5,y=-1/5,n为自然数,求x^3n*x^3*y^2n*y^4的值如题.._数学解答

x=5,y=-1/5,n为自然数,求x^3n*x^3*y^2n*y^4的值
如题..

人气：415 ℃　时间：2020-05-08 22:55:25

解答

x^3n*x^3*y^2n*y^4
=x^(3n+3)*y^(2n+4)
=(x^3)^(n+1)*y^(2n+2)*y^2
=(x^3)^(n+1)*(y^2)^(n+1)*y^2
=(x^3*y^2)^(n+1)*y^2
=(125*1/25)^(n+1)*(1/25)
=5^(n+1)*1/25
=5^n*5*1/25
=(5^n)/5