已知3^m=5,3^n=2
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∵2n-m=5∴m+5=2n5(m+5)²+6n-3m-60=5×4n²+6n-3(m+5)-45=20n²+6n-6n-45=20n²-45明教为您解答,如若
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
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
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/(m+n)+m/(m-n)-n²/((m²-n²)=(m²-mn+m²+mn-n²)/(m²-n²)=(2m&sup
(3m--2n)--(2m--3n)=3m-2n-2m+3n=m+n=0再问:等于0吗再答:是的,等于0,如果有帮助,请给个好评吧再答:给个好评哈,谢谢!再问:咋给再答:追问附近找找哈,再问:???再
"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
合并多项式,(m/m+n)+(m/m-n)-(m^2/m^2-n^2)=2m^2/(m^2-n^2)-(m^2/m^2-n^2)=m^2/(m^2-n^2)将m=5n/3代入得,(5n/3)^2/[(
[(3m+2n)(3m-2n)-(m+2n)(5m-2n)]÷(1/3)m=[9m²-4n²-5m²+2mn-10mn+4n²]÷(1/3)m=[4m²
199-m-n>=0,m+n
由m/n=5/3,得m=5n/3原式={(m-n)/(m-n)*(m+n)+(m+n)/(m-n)*(m+n)}*n-n平方/(m-n)*(m+n)=2mn/(m平方-n平方)-n平方/(m平方-n平
∵mn+3m+5n=70∴(m+5)(n+3)=85∵85=5X17=1X85∵m.n是正整数∴(m+5)和(n+3)只能取5和17∵m+5>5,n+3>3∴m+5=17,n+3=5∴m=12,n=2
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
这种题最好能够把括号加上,不然很难分辨题目是什么合并多项式,(m/m+n)+(m/m-n)-(m^2/m^2-n^2)=2m^2/(m^2-n^2)-(m^2/m^2-n^2)=m^2/(m^2-n^
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/M+2n的绝对值=32m-n/M+2n=3或-32(2m-n)/m+2n减2m-n/m+2n减3=(2m-n/m+2n)-3=0或-6
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
由m/n=5/3,得m=5n/3原式={(m-n)/(m-n)*(m+n)+(m+n)/(m-n)*(m+n)}*n-n平方/(m-n)*(m+n)=2mn/(m平方-n平方)-n平方/(m平方-n平
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