(25．4-4m)÷4解-搜答案-搜搜做题作业网

[(m-2)/(m+2)+4m/(m²-4)]÷[1/(4-m²)]=[(m-2)²+4m]/(m+2)(m-2)×(2-m)(2+m)=-(m²-4m+4+4

先合并同类项

原式=m-{3n-4m+[m-5m+5n+m]}=m-{3n-4m+5n-3m}=m-3n+4m-5n+3m=8m-8n．故答案为：8m-8n．

(m+2/m²-2m)-（m-1/m²-4m+4）÷（m-1/m）=(m+2)/m(m-2)-(m-1)/(m-2)^2*m/(m-1)=(m+2)/m(m-2)-m/(m-2)^

(m+3m+5m+.+2009m)-(2m+4m+6m+.+2008m)=2009m

化简(m*m-2m)/(4-m*m)

=m(m-2)/(2-m)(2+m)=-m/(m+2)

3m-2m=m5m-4m=m……2009m-2008m=m相当于m+m+m+……m总共2009答案2009m

(4m²-4m+1)/(m+1)÷(4m²-1)/(m²-1)=(2m-1)²/(m+1)×(m+1)(m-1)/(2m+1)(2m-1)=(2m-1)(m-1

按照系数大小重新排序m-2m+3m-4m……+2009m-2010m+2010m每两项组成一组结果为1005个-m加2010m=1005m

=（m-2）²/(m+1)(m-1)*(m+1)/m-2+2/m-1=（m-2）/(m-1)+2/m-1=m/m-1=2

(m+3m+5m+...+2009m)-(2m+4m+...+2008m)=m+(3-2)m+(5-4)m+……+(2009-2008)m=m+m+m+……+m=1005m嗯再问：有没有可能的-100

m-2m-3m+4m-5m-6m+7m.2004m=(1-2-3+4-5-6+7-8-9+...2002-2003-2004)m=(-4-7-10-..-2005)m=-(4+7+10+..+2005

(m+3m+5m+...+2009m)-(2m+4m+6m+...+2010m)=2010m/2=1005m

(m+3m+5m+...+2013m)-(2m+4m+6m+...+2012m)=(2013m-2012m)+(2011m-2010m)+...+(5m-4m)+(3m-2m)+m=m+m+...+m

（m2013m）*1007/2－（2m2012m）*1006/2=1007m

原式=m+3m+5m+.2014m-2m-4m-6m-.2013m=m-2m+3m-4m+5m-6m=m+(3m-2m)+(5m-4m)+(2014m-2013m)=m+m+.1007个m=1007m

=m+(3m-2m)+(5m-4m)+……+(2015m-2014m)=m+1007m=1008m

有规律的,自己仔细看看～

原式=-(m+4)(m-4)/(m+4)²×2(m+4)/(m-4)×(m-2)/(m+2)=-(m-4)/(m+4)×2(m+4)/(m-4)×(m-2)/(m+2)=-2×(m-2)/(

m2（m+4）+2m（m2-1）-3m（m2+m-1）=m3+4m2+2m3-2m-3m3-3m2+3m，=m2+m，当m=25时，原式=425+25=1425．