若 3 m =6 , 9 n =2 ,则 3 2m−4n 1 =

既然是相反数,就是m和n两者加起来为0的,例如-1和1,-2和2就互为相反数,所以这里答案是0咯

解|m-n|+(m+2n-3)²=0∴m-n=0①m+2n-3=0②∴m=n将m=n代入②得：3n-3=0∴n=m=1∴m+n=1+1=2

n/m=3/4m/(m+n)=1/[1+n/m]=4/7n/(m-n)=1/[m/n-1]=3m^2/(m^2-n^2)=1/[1-n^2/m^2]=16/7所以原式=9/7当然,你可以通分来算,也能

m-{n-2m+[3m-(6m+3n)+5n]}=m-n+2m-[3m-(6m+3n)+5n]=3m-n-3m+(6m+3n)-5n=-6n+6m+3n=6m-3n

m+2=n-3m+2-4=n-3-4m-4=n-3-4-2=n-9

(2x^3m*x^2n)/(x^m*x^n)^2*x^n=（2*6^3*9^2)/(6*9)^2*9=4/3

m/(m+n)-m/(m-n)-2n^2/(m^2-n^2)=m(m-n)/(m^2-n^2)-m(m+n)/(m^2-n^2)-2n^2/(m^2-n^2)=(m^2-mn-m^2-mn-2n^2)

将m+n看作x,那么算式变成了x的方-6x+9,那么用平方公式一算,就是（x-3）的方,所以答案不就是（m+n-3）的方嘛!至于-3嘛,-3乘以2不就是-6嘛!希望采纳,谢谢!

(m-2n)/(2m+n)=3m-2n=6m+3n5m+5n=0m=-n[3(m-2n)/(2m+n)-(m-2n)/[2(2m+n)]-{9(m-2n)/[4(2m+n)]}=[3(-n-2n)/(

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

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)

m²+²-6n+4m+13=0(m²+4m+4)+(n²-6n+9)=0(m+2)²+(n-3)²=0两非负数和为0,故(m+2)²

因为m、n互为相反数,所以m+n=0(5m-3n)-(2m-6n)=5m-3n-2m+6n=3m+3n=3(m+n)=0

x²+5y²-4xy+2y+1=0x²+4y²-4xy+y²+2y+1=0(x-2y)^2+(y+1)^2=0∵(x-2y)^2>=0,(y+1)^2

2m-3n/3m+5n=3则：3m+5n/2m-3n=1/3(4m-6n/18m+30n)+9m+15n/4m-6n=2（2m+3n)/6(3m+5n)+3(3m+5n)/2(2m-3n)=1/3*3

由m×m+n×n-6n+4m+13=0得(m×m+4m+4)+(n×n-6n+9)=0得(m+2)^2+(n-3)^2=0得m=-2n=3所以m^2-n^2=4-9=-5

(2m-n)/(m+2n)=5∴(m+2n)/(2m-n)=1/5∴3(2m-n）/m+2n-m+2n/2m-n+3=3×(2m-n)/(m+2n)+(m+2n)/(2m-n)+3=15+1/5+3=

题错了应该是m^3-m^2n-mn^2+n^3=m^3+n^3-mn(m+n)=m^3+n^3-0=(m+n)(m^2-mn+n^2)=0

6m（m-n）-4n（n-m）=6m3+4n（m-n）=18m+4×3n=6×（3m+2n）=6×2=12

1/2（m-n)+1/4(m+n)+m+n/3-m-n/6=1/2m-1/2n+1/4m+1/4n+m+n/3-m-n/6=(1/2m+1/4m+m-m)+[(-1/2n+1/4n)+（n/3-n/6