m*n=5m-3n那么(5分之4*4分之3)*2又3分之2

已知-m+2n=5,那么（m-2n）²+6n-3m-60=__-20_____.-m+2n=52n-m=5m-2n=-56n-3m=3（2n-m）（m-2n）²+6n-3m-60=

m/5+n/2=2(1)2m-3n=4(2)去分母,(1)×10得：2m+5n=20(3)消去m,(3)-(2)得：5n-(-3n)=20-48n=16n=2代入(2)得：2m-3×2=42m-6=4

(m+n分之m)+(m-n分之m)-(m方-n方,分之n方)=m(m-n)/(m^2-n^2)+m(m+n)/(m^2-n^2)-n^2/(m^2-n^2)=(m^2-mn+m^2+mn-n^2)/(

(n/m)^(-1)=m/n=5/35n=3mn=0.6mn²=0.36m²所以原式=[m(m-n)+m(m+n)-n²]/(m+n)(m-n)=(2m²-n&

m/(m-n)-n/(m+n)+2mn/(m²-n²)=[m(m+n)-n(m-n)+2mn]/(m²-n²)=(m²+n²)/(m

答：mn/(m+n)=2分子分母同除以mn得：1/(1/n+1/m)=21/m+1/n=1/2(3m-5mn+3n)/(-m+3mn-n)分子分母同除以mn得：=(3/n-5+3/m)/(-1/n+3

∵n分之m=3分之5∴n=3m/5(m+n分之m)+(m-n分之m)-(m方-n方,分之n方)=m/(m+n)+m/(m-n)-n²/(m²-n²)=(m²-m

3/4m+2n=1/3m+5n=2/5m+n+2上面的方程可以变成二元一次方程组,即3/4m+2n=1/3m+5n1/3m+5n=2/5m+n+2整理,得5m-36n=0①m-60n=30②②×5,得

m/(m+n)+m/(m-n)-n²/((m²-n²)=(m²-mn+m²+mn-n²)/(m²-n²)=(2m&sup

原式=m/(m+n)+m/(m-n)-n^2/(m^2-n^2)=(m^2-mn+m^2+mn)/(m^2-n^2)-n^2/(m^2-n^2)=(2m^2-n^2)/(m^2-n^2)因为m/n=5

应该是-m+2n=5吧=5(-m+2n)^2+3(-m+2n)-60=5*5^2+3*5-60=125+15-60=80

(1)∵m/n=5/3,可得：3*m=5*n,m=(5/3)*n;∴(m+n)/(m-n)=((8/3)*n)/((2/3)*n)=4.（2）我算不出框重,条件不足

把题目传上来,要不然自己去化简,或者可以代掉一个字母,根据n分之m=3分之5,用一个字母表示另一个,最后肯定可以约掉再问：已知n分之m=3分之5，求（m+b分之m）+（m-b分之m）-（m平方-n平方

答：-m+2n=5(m-2n)²+6n-3m-60=(-m+2n)²+3(2n-m)-60=5²+3*5-60=25+15-60=-20

m/n=3/5则n/m=5/3m/(m+n)+m/(m-n)-[n^2/(m^2-n^2)]=[m(m-n)+m(m+n)-n^2]/(m^2-n^2)=(2m^2-n^2)/(m^2-n^2)=1+

已知m=5n,则原式=(5n/(5n+n))+(5n/(5n-n))-(n^2)/(((5n)^3)-n^2)=(5/6)+(5/4)-[1/(125n-1)]=(25/12)-[1/(125n-1)

5(m-2n)^2+6n-3m-60=5(-m+2n)^2+3(-m+2n)-60=5*5^2+3*5-60=125+15-60=80

解m/n=1/5∴n=5m∴(m+n)/(m-n)=(m+5m)/(m-5m)=6m/(-4m)=-3/2

2m-3n=2m+2n=14两式相减可得m-5n=-12所以m-5n+3=-12+3=-9

2m+n分之m-2n=3所以原式=(3+1/3-5/4)[2m+n分之m-2n=3]=25/12×3=25/4