用加减消元法解方程{(m+n)/3-(m-n)/4=1 2m+(3/4)n=-29
4m-3n+1=0 2m-6n=7 用加减消元法解
{(m+n/2)+(m-2/3)=6,4(m+n)-5(m-n)=2 用加减消元法解二元一次方程急求解答谢谢!
3m+5n=12① 3m+6n=-11②用加减消元法解方程.
用加减消元法解方程组 5(m-1)=2(n+3) { 2(m+1)=3(n-3)
2分之m+n-3分之m-n=1,3分之m+n-4分之m-n=-1
1,已知3m=4n,则m/m+n +n/m-n -m^2/m^2-n^2=
如果m-3n+4=0 求(m-3n)^2+7m^3-3(2m^3n-m^2n-1)+3(m^3+2m^3n-m^2n+n
如果m和n互为相反数,则m+2m+3m+4m+5m+5n+4n+3n+2n+n=
如果m,n互为相反数,求m+2m+3m+4m+5m+5n+4n+3n+2n+n=多少
[(m-n)^2*(m-n)^3]^2/(m-n)^4
(m-n)2(n-m)3(n-m)4化简
先化简,在求值(m+n)(m-n)(-m^2-n^2)-(-2m+n)(-2m-n)(4m^2+n^2) 其中m=1,n