m=2*3*5 n=2*3*7他们的
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5m+3n=85m-3n=2两式子相加得10m=10∴m=15m+3n=85+3n=8n=1m+n=1+1=2
此题先通分,通过计算可得m^2+m^2-n^2/m^2-n^2=1+m^2/m^2-n^2=1+25/16=41/16
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
m/(m+n)+m/(m-n)-n^2/(m^2-n^2)=m(m-n)/(m+n)(m-n)+m(m+n)/(m-n)(m+n)-n^2/(m^2-n^2)=(m(m-n)+m(m+n))/(m^2
3m+2n=7(1)3m-n=5(2)(1)-(2)得:2n+n=7-5=2;n=2/3;带入(1)得:3m+4/3=7;3m=17/3;m=17/9;如果本题有什么不明白可以追问,
已知,m=5又4/7,n=4又3/7,可得:m+n=10,m-n=1又1/7=8/7.[-3又1/2(m+n)]^3×(m-n)×[-2(m+n)(m-n)]^2=(-7/2)^3×(m+n)^3×(
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/(m+n)+m/(m-n)-n^2/(m^2-n^2)=(m/n)/(1+m/n)+(m/n)/(-1+m/n)-1/((m/n)^2-1)=(5/3)/(1+5/3)+(5/3)/(5/3-1
[(3m+2n)(3m-2n)-(m+2n)(5m-2n)]÷(1/3)m=[9m²-4n²-5m²+2mn-10mn+4n²]÷(1/3)m=[4m²
答:(5m+3n)^2-(m-3n)(25m-3n),其中m=1,n=2=25m²+30mn+9n²-(25m²-3mn-75mn+9n²)=30mn+78mn
(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)
2m-n=-75m-3n=2解6m-3n=-215m-3n=2代入6m-3n-(5m-3n)=-23m=-23-64-n=-7n=-57m=-23n=-57如有问题请追问,如果对我得回答满意这是你对我
(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=
(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=
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