y=[2m 4]x [3-n]m,n是什么数,函数图像经过原点

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y=[2m 4]x [3-n]m,n是什么数,函数图像经过原点
设集合M={x/x=3m+1,m属于Z},N={y/y=3n+2,n属于Z},若x属于M,y属于N,则xy与集合M,N的

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

x2+y2-2(m+3)+2(1-4m2)y+16m4+9=0是一个圆,求半径r取值范围

[x-(m+3)]^2+[y+(1-4m^2)]^2=-16m^4-9+(m+3)^2+(1-4m^2)^2所以r^2=-16m^4-9+(x+3)^2+(1-4m^2)^2>0-16m^4-9+(m

已知,单项式M,N满足2x(M+3x)=6x²y²+N,求M,N

2x(M+3x)=6x²y²+N2xM+6x²=6x²y²+N所以N=6x²,且2xM=6x²y²,即M=3xy&sup

设方程x2+y2-2(m+3)x-2(1-4m2)y+16m4+9=0.若该方程表示一个圆,求m的取值范围.

圆的方程化为[x-(m+3)]2+[y-(1-4m2)]2=1+6m-7m2,则有1+6m-7m2>0,解得-17<m<1,故m的取值范围是(-17,1).

已知方程x2+y2-2(m+3)x+2(1-4m2)y+16m4+9=0表示一个圆.

(1)由方程x2+y2-2(m+3)x+2(1-4m2)y+16m4+9=0变形得:[x-(m+3)]2+[y+(1-4m2)]2=-7m2+6m+1,当且仅当-7m2+6m+1>0,即7m2-6m-

已知抛物线y=x-3x+1经过(m,0)求m4-21m+10

y=x-3x+1经过点(m,0),代入得到m^2-3m+1=0m^2=3m-1m4-21m+10=(3m-1)^2-21m+10=9m^2-6m+1-21m+10=9m^2-27m+11=9(3m-1

已知,单项式M,N满足2x(M+3x)=6x^2y^2+N,求M,N.

解析2*(M+3x)=2*m+6x=6x²y²+N两边比较2*m=6x²y²m=3x²y²N=6x再问:我没看明白。请问这几个算式都是为什么

已知单项式M,N满足3x(M-5x)=6x^2y^3+N,求M,N

3x(M-5x)=6x^2y^3+N,3x*M-15x^2=6x^2y^3+NM=2xy^3N=-15x^2

已知单项式M、N满足2x(M+3x)=6x^2y^2+N,求M、N.

由于2x(M+3x)=6x^2y^2+N,所以2xM+6x^2=6x^2y^2+N比较可知2xM=6x^2y^2,6x^2=N所以M=3xy^2,N=6x^2

若关于x的分式方程2x−4=3+m4−x有增根,则m的值是(  )

方程两边都乘(x-4),得2=3(x-4)-m∵当最简公分母x-4=0时,方程有增根,∴把x-4=0代入整式方程,∴m=-2.故选A.

计算:1.[(m+n)(n-m)-(m+n)^2+3m(m+n)]除以(-2M),其中m=-4,n=2 2.(x+y)(

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

若[-3(x+y)^m*(x-y)^2n]^2*[-(x+y)^2]=-a(x+y)^10*(x-y)^12-n,求m,

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

集合M={X=3m+1,m∈N*},N={y/y=5n+2,n∈N*},则M∩N=?

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.

把下列各式因式分解(2m+3n)(2m-n)-4n(2m-n)(x+y)^2(x-y)+(x+y)(y-x)^24a(x

(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+

已知方程x2+y2-2(m+3)x+-2(1-4m2)y+16m4+9=0表示一个圆,求圆心的轨迹方程

[x-(m+3)]^2+[y-(1-4m^2)]^2=-(7m^2-6m-1)=-(7m+1)(m-1)>0-1/7再问:半径是多少再答:根号下[-(7m+1)(m-1)]

设方程x2+y2-2(m+3)x+2(1-4m2)y+16m4+9=0.

(1)由方程x2+y2-2(m+3)x+2(1-4m2)y+16m4+9=0变形得:[x-(m+3)]2+[y+(1-4m2)]2=-7m2+6m+1,要使方程表示圆,则需要-7m2+6m+1>0;如

已知x^3m=8,y^3n=27,求(x^2m)^3+(y^n)^3-x^2m*y^n

x^3m=8,y^3n=27,所以x^m=2,y^n=3,所以(x^2m)^3+(y^n)^3-x^2m*y^n=(2^2)^3+3^3-2^2*3=64+9-12=61

2x^6n-x^4n×y^3m-y^9m,已知x^2n=3,y^3n=2,求值

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

若3x^m+1y^2与x^3y^n的积是单项式3x^5y^5则n^m+m^n=

.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