m-2份之N=2 2M+3N=12用代如消元
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m-2份之N=2 2M+3N=12用代如消元
简单的初一数学题(2元1次方程组){5份之M-2份之N=2 2M+3N=4+-
(m-n)^3(n-m)^2(m-n)=_____
(3m-n)(m-2n)=()
(m-2n/n-m)-(n/m-n)=
已知3m=2n,则m/(m+n)+n/(m-n)-n^2/(m^2-n^2)=?
已知3m=4n,则m/m+n+n/m-n-m^2/m^2-n^2=
m/n=5/3,求(m/m+n)+(m/m+n)-(n^2/m^2-n^2)
已知m/n=5、,求(m/(m+n))+(m/(m-n))-(n^2/(m^3-n^2))
2m+3n+2m+3n+m+2m+3n+n=______
(m-2n)(2n+m)-(-3m-4n)(4n-3m)=
.4(m+n)^2*(m+n)^3-7(m+n)(m+n)^4+5(m+n)^5=
(m-n)^2*(m-n)^4=?
m-{n-2m+[3m-(6m+3n)+5n]}=
(m-n)(m+n)+(m+n)²-3=2m²
(2m+n)(2m-n)+n(2m+n)-8m^2n^3/(2n^3) m=-2分之1,n=2013
已知|2m-n/m+2n|=3,求2(2m-n)/m+2n-2m-n/m+2n-3的值
已知|2m -n/m+2n|=3求2(2m-n)/m+2n-2m-n/m+2n-3的值