3n-2m-1=3m-2n运用等式性质,比较m与n大小
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/15 05:31:38
(m-n)2(n-m)7
一项一项通分,分母用平方差公式合并,分子进行加减,很简单就能算出来.别图省事,只要动手就能算出来的.
(m+n)^3-(n+m)(m-n)^2=(m+n)[(m+n)^2-(m-n)^2]=(m+n)[(m+n+m-n)(m+n-m+n)]=(m+n)(2m+2n)=2(m+n)^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
[(3m+2n)(3m-2n)-(m+2n)(5m-2n)]÷(1/3)m=[9m²-4n²-5m²+2mn-10mn+4n²]÷(1/3)m=[4m²
原式=m(3m-2n+1)-n(3m-2n+1)=3m²-2mn+m-3mn+2n²-n=3m²-5mn+2n²+m-n
(n-m)^3×(m-n)^2-(m-n)^5=-(m-n)^3*(m-n)^2-(m-n)^5=-(m-n)^5-(m-n)^5=-2(m-n)^5
答:(5m+3n)^2-(m-3n)(25m-3n),其中m=1,n=2=25m²+30mn+9n²-(25m²-3mn-75mn+9n²)=30mn+78mn
设m-n为a(a^2*a^3)^2/a^4=a^6即(m-n)^6
(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)/(
解原式=-(m-n)³(m-n)²-(m-n)^5=-(m-n)^5-(m-n)^5=-2(m-n)^5
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
C(m+1,n)=C(m,n-1)+C(m+1,n-1)这个式子可以直接验证,也可以算两次得证.然后递推C(m+1,n)=C(m,n-1)+C(m+1,n-1)=C(m,n-1)+C(m,n-2)+C
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)
4.5*5=4.5+(4.5+1)+(4.5+2)+(4.5+3)+(4.5+4)+4.5+5)=4.5x6+1+2+3+4+5=27+15=42m*8=37.8m*n=m+(m+1)+(m+2)+(
原式=4m²-n²+mn+2n²-4m²-4mn-n²=-3mn
根据原式可知:m-3n=1,且2m+n-15=1,将m-3n=1移项后为m=1+3n,将其代入2m+n-15=1中:2×(1+3n)+n-15=17n=14n=2m-3×2=1m=7
既然m-3n+4=0那么m-3n肯定等于-4,再把-4带进去算就是了
3m(m-n)-2n(m-n)²=(m-n)(3m-2mn+2n²)