(m 2)2004 (2m-5 7)-搜答案-神马作文网

m²+m≠0m(m+1)≠0m≠0且m≠-1m²-m=2m²-m-2=0(m-2)(m+1)=0m=2或m=-1再问：不是啊，答案只有2再答：我以为是2个问-_-m

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

m2-3m+2 _________m2-m

(m-2)/m

m2-3m+2 m-2

(m-1)*(m-2)/m(m-1)同时去掉m-1结论就对了=(m-2)/m

解题思路：将原代数式经过去括号后合并同类项，化简后代入数值计算即可。解题过程：

(m-2)(m2+2m+4)-2m(m2-3)=m^3-8-2m^3+6m=-m^3+6m-8其中m=2.=-8+12-8=-4

1/(m2-m)+(m-5)/(2m2-2)

1,=1/m(m-1)+(m-5)/2(m+1)(m-1)=2(m+1)/2(m+1)(m-1)+(m2-5m)/2(m+1)(m-1)=(2m+2+m2-5m)/2(m+1)(m-1)=(m2-3m

计算 12/m2-9+2/m-3

12/(m²-9)+2/(m-3)=12/(m²-9)+2(m+3)/(m²-9)=2(m+9)/(m²-9)

多项式-2m2+3m-12

多项式-2m2+3m-12的各项系数之积为：-2×3×（-12）=3．故答案为：3．

设m2次-2m+3为A,用换元法原式=A-8A+12用十字相乘=(A-2)(A-6)=(m2次-2m+3-2)(m2次-2m+3-6)=(m2次-2m+1)（m2次-2m-3）前面公式法后面十字相乘=

(m^2-m)/(m^2-3m+2)=m(m-1)/(m-2)(m-1)=m/(m-2)

(m+n)2-4(m2-n2)+4(m-n)2

(m+n)2-4(m2-n2)+4(m-n)2=（m+n)^2-2*2(m+n)(m-n)+(2(m-n))^2=(m+n-2(m-n))^2=(m+n-2m+2n)^2=(3n-m)^2

∵m²+2m+1＝0∴m=-1∴m³+2m²+3m=-2

-8m2-[4m-2m2-（3m-m2-7）-8]=-8m2-[4m-2m2-3m+m2+7-8]=-8m2-（-m2+m-1）=-8m2+m2-m+1=-7m2-m+1

2m2-3m-4+（m2+2m）=2m2-3m-4+m2+2m=3m2-m-4．故答案为：3m2-m-4．

解析：（1）已知m²+m-1=0,那么：m²+m=1所以：m³+2m²+2001=m³+m²+m²+2001=m(m²+

(m^2十4m)^2十8(m^2十4m)十16令t=m^2十4m,则(m^2十4m)^2十8(m^2十4m)十16=t^2+8t+16=(t+4)^2=(m^2十4m+4)^2=(m+2)^4

如果m-m/1则m2次加m等于（0),2m2次加2m-1等于（0)

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．