若X^4-3|m| y^ |n|-2=2013是关于x,y的二元一次方程,且-搜答案-百度作业网

先假设x=1y=1,得2m+n=13m+2n=1,算出m=1n=-1

M集合肯定有一个元素2,所以log(2x)4=2,x=正负1,-1(舍).所以M集合{3,2},N集合{1,2},所以答案是{1,2,3}

m+2n=3①1=3m-2n②联立,解得m=1,n=1

=0

x*y=(x+1)(y+4)-(x+2)(y+3),有：（m+1)*(n-1)=[（m+1)+1）][(n-1)+4]-[(m+1)+2][(n-1)+3]=(m+2)(n+3)-(m+3)(n+2)

m²n-2m+n

x=m^3(m-n),y=n^3(m-n)x-y=(m^3-n^3)(m-n)=(m-n)(m^2+mn+n^2)(m-n)=(m-n)^2*(m^2+mn+n^2)因为m≠n,所以(m-n)^2>0

1.[(m+n)(n-m)-(m+n)^2+3m(m+n)]÷（-2m)=(m+n)[(n-m)-(m+n)+3m]÷（-2m)=(m+n)(n-m-m-n+3m)÷（-2m)=m(m+n)÷(-2m

由题意：m+n-2=1m-4-n=1两式相加得：2m-6=22m=8m=4所以：n=1-m+2=-1选择C

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

(2m+3n)(2m-n)-4n(2m-n)=(2m-n)(2m+3n-4n)=(2m-n)(2m-n)=(2m-n)^2(x+y)^2(x-y)+(x+y)(y-x)^2=(x+y)(x-y)(x+

（3m-4n)(3m+4n)-(2m-2n)(2m+n)=9m²-16n²-4m²+2mn+2n²=5m²+2mn-14n²(x+5y)&#

x-y=m^4-m^3*n-n^3*m+n^4=m(m^3-n^3)-n(m^3-n^3)=(m-n)(m^3-n^3)因为,x!=y,所以,不可能有,m=n的情况.由上式易见,无论m>n还是m0即,

(m+1)&(n-1),即x=m+1,y=n-1所以(m+1)&(n-1)=(m+1+1)(n-1+4)-(m+1+2)(n-1+3)=(m+2)(n+3)-(m+3)(n+2)=mn+3m-2n+6

由题目知：2m-n=1(1)3n-m=1(2)由（1）得,n=2m-1(3)把（3）代入（2）得3（2m-1)-m=16m-3-m-1=05m-4=0m=4/5n=2*4/5-1n=3/5

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

若(-5x^m+1×y^2-1)×(2x^n×y^m)=-10x^4y^4,试求m-n的值.m+1+n=4；2-1+m=4；m=3；n=0；∴m-n=3-0=3；x^3-2x[2x^2-3(x^2-2

原式=x^6m+(y^3n)^2=(x^3m)^2+(y^3n)^2=4^2+5^2=41

(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