若x^3n=2,y^2n=3,求(xy)^6n-搜答案-智慧作业帮

xy属于N因为x属于M,y属于N可设x=3m+1,y=3n+2则xy=(3m+1)*(3n+2)=9mn+6m+3n+2=3(3mn+2m+n)+2由于m,n均属于Z,则3mn+2m+n属于Z显然xy

x^(6n)+y^(4n)=[x^(3n)]^2+[y^(2n)]^2=5×5+4×4=25+16=41

x=√(n+3)-√(n+1),y=√(n+2)-√n显然x>0,y>01/x=1/[√(x+3)-√(n+1)]=[√(n+3)+√(n+1)]/[(n+3)-(n+1)]（分母有理化）=[√(n+

原式=x^([n+(n-1)+……+2+1]y^(1+2+3+……+n)=x^ay^a再问：好快啊再答：采纳吧再问：可以再详细一点吗再答：采纳吧再问：我很怀疑你也是百度的呢

x^6n+y^6n*x^4n=(x^3n)^2+(y^2n)^3*x^3n*x^n=4+54x^n

（xy)^2n=x^2n*y^2n=(x^n)^2*(y^n)^2=5^2*3^2=25*9=225(x^2y^3)^n=x^2n*y^3n==(x^n)^2*(y^n)^3=25*27=675

9（x+y)^(2m)*(x-y)^(4n)*[-(x+y)^2]=-9(x+y)^(2m+2)*(x-y)^(4n)∴a=92m+2=104n=12-n∴m=4n=12/5

3m=5n+13m-6n=-n+1n=3(2n-m)+1=3(2n-m+1)-2所以可令n=3k+1或者3k-2.

（y^n)^3=y^3n=3所以原式=x^6n*y^6n=(x^2n)^3*(y^3n)^2=2^3*3^2=72

把b=xy+xz+yz,c=xyz代入,可得恒等式,即证毕

x^2n=2,x^6n=2^3=8,(y^n)^3=3,y^6n=3^2=9则(xy)^6n=x^6n*y^6n=8*9=72

结果是0

x^2n=3x^6n=(x^2n)^3=3^3=27x^4n=(x^2n)^2=3^2=9y^9n=(y^3n)^3=2^3=82x^6n-x^4n×y^3m-y^9m=2*27-9*2-8=28

.3x^m+1y^2与x^3y^n的积是3x^(m+4)y^(2+n),即是单项式3x^5y^5则有m+4=5,n+2=5m=1,n=3n^m+m^n=1+3=4再问：我也是这个答案，但是孩子说老师说

=X^4n*y^2n=(X^n)^4*(y^n)^2=5^4*3^2=625*9=5625

(x^2m)^3+(y^n)^3-x^2m*y^n*x^4m*y^2n=x^6m+y^3n-x^6my^3n=(x^3m)^2+y^3n-(x^3m)^2*y^3n=4^2+5-4^2*5=-59

原式=x^4ny^2n=(x^n)^4(y^n)^2=2^4×3^2=144

（x^2y)^2n=x^4n×y^2n=(x^n)^4×(y^n)^2=2^4×3^2=16×9=144

(x^2y^2)^2n=[(xy)^2]^2n=(xy)^4n=[(xy)^n]^4=(x^n*y^n)^4=(5*3)^4=50625