c语言main(){intx=3,y=4,n;n=x-y++;n=3-5=-2n+=x*y;n=n+

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c语言main(){intx=3,y=4,n;n=x-y++;n=3-5=-2n+=x*y;n=n+
已知x^n=5,y^n=3,求(x^2y)^2n

已知x^n=5,y^n=3,求(x^2y)^2n已知x^n=5,y^n=3,求(x^2y)^2n已知x^n=5,y^n=3,求(x^2y)^2n原式=x^4ny^2n=(x^n)^4(y^n)^2=5

C语言 main() {int x,y; x=11/3;y=5; printf("%%d,%%%d ...

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计算:(x^2n-2x^ny^n+y^2n)/(x^n-y^n)(n为正整数)=

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若X^n=2,y^n=3,(x^2y^3)^n=

若X^n=2,y^n=3,(x^2y^3)^n=若X^n=2,y^n=3,(x^2y^3)^n=若X^n=2,y^n=3,(x^2y^3)^n=108

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若X^n=2,y^n=3,则(xy)^n=(),(x^2y^3)^n=()若X^n=2,y^n=3,则(xy)^n=(),(x^2y^3)^n=()若X^n=2,y^n=3,则(xy)^n=(),(x

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#include#includevoid main(){double y,x,n;n=1;printf(Enter x(x>1) );scanf(%f,&x);for(y=1;y>=0.000001;n++){ y=y+(1/pow(x,n));printf(%f ,y);}}

#include#includevoidmain(){doubley,x,n;n=1;printf("Enterx(x>1)\n");scanf("%f",&x);for(y=1;y>=0.00000

x^2n=3,y^3n=4 (xy)^6n=?

x^2n=3,y^3n=4(xy)^6n=?x^2n=3,y^3n=4(xy)^6n=?x^2n=3,y^3n=4(xy)^6n=?x^2n=3,y^3n=4(xy)^6n=(x^2n)^3*(y^3

已知x^n=5,y^n=3,求(xy)^2n的值

已知x^n=5,y^n=3,求(xy)^2n的值已知x^n=5,y^n=3,求(xy)^2n的值已知x^n=5,y^n=3,求(xy)^2n的值(xy)^2n=[(xy)^n]²=(x^ny