[3m-2n+1][3m+2n-1]
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[3m-2n+1][3m+2n-1][3m-2n+1][3m+2n-1][3m-2n+1][3m+2n-1]=[3m-(2n-1)][3m+2n-1]=9m²-(2n-1)²原式=
[3m-2n+1][3m+2n-1]
[3m-2n+1][3m+2n-1]
[3m-2n+1][3m+2n-1]
=[3m-(2n-1)][3m+2n-1]
=9m²-(2n-1)²
原式=[3m-(2n-1)][3m+(2n-1)]
=(3m)²-(2n-1)²
=9m²-4n²+4n-1
(1),(-m-n)(-m+n) (2),(-m+n)(m-n)
1/m-n,2/n^2-m^2,3n/2(m+n)通分
(2m+n)(2m-n)+n(2m+n)-8m^2n^3/(2n^3) m=-2分之1,n=2013
若2m=3n,求分式(1+m/n-n/n-m)除以(1-m/n-n/n+m)的值
若2m=3n,求分式(1+m/n-n/n-m)除以(1-m/n-n/n+m)的值
(m-2n)^2+(3m-n)(2m+2n)-(2m+n)(2m-n) 已知m+n=4 mn=1
1/(m-n)-1/(m+n)-2n/(m^2+n^2)-4n^3/m^4+n^4)-8n^7/(m^8+n^8)-16n^15/(m^16+n^16)+32n^31/(n32-m32)
m^2=n+2,n^2=m+2(m不等于n),求m^3-2mn+n^3的值因为m^2=n+2,n^2=m+2 所以 m^2-n^2=n-m 即 (m-n)(m+n)=n-m m+n=1为什么得m+n=1,求详解.
[3m-2n+1][3m+2n-1]
(3m+2n-1)(1+3m-2n)
(3m-2n)(3m+2n+1)求解
(1/3m+0.75n)(0.8m-2/3n)
(m-n)*(3m-2n+1)=?
解方程组:(m+n)-2(m-n)=1 3(m+n)+2(m-n)=3 (m+n)-2(m-n)=1 3(m+n)+2(m-n)=3 是这两个方程
1若m=n,则|m|=|n|;2若m=—n,则|m|=|n|;3若|m|=|n|,则m=—n;4若|m|=|n|,则m=n哪个对
先化简,再求值;(2m+n)(2m-n)+n(m+2n)-(2m+n)的平方,其中m=-3/1,n=2011
数学题[(m+3n)(m-3n)+(2n-m^2)+5n^2(1-m)-(2m^2-m^2n)÷(-1/2mn)
{(2/m+n)+(3/m-n)=3/2 {(1/m+n)-(9/m-n)=-1