jduizhim-1 m-2化简-搜答案-拍题作业网

|2^m-2^[8/(2m+1)]|/|2^(-m)-2^[-8/(2m+1)]|=|2^m-2^[8/(2m+1)]|/|{1/2^m-1/2^[8/(2m+1)]|=|2^m-2^[8/(2m+1

当m＜0时，m-1＜0，m-2＜0，原式=-m+1-m+2-m=3-3m；当0＜m＜1时，m-1＜0，m-2＜0，原式=m+1-m+2-m=3-m；当1≤m＜2时，m-1≥0，m-2＜0，原式=m+m

化简|m+2|－|－m-1|

当m=

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

∵m>4∴m-4>0,7-2m0,m+1>0,m-3>0∴(m+1)(m-3)>0∴原式=|m-4|+|7-2m|+|(m-1)²|-|(m+1)(m-3)|=m-4+2m-7+(m-1)&

4m+8

由原题可知m-2=|m-2|∴m-2≥0即m≥2∴1-m

化简(m*m-2m)/(4-m*m)

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

2［（m－1）m＋m（m＋1）］［（m－1）m－m（m＋1）］=2(m²-m+m²+m)(m²-m-m²-m)=2×2m²×(-2m)=-8m

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

化简|2m+1|+|m-2|

三种情况.1.当m

由m0,2m+1

(m-1)+2(m²-3m)-(2-m+5m²)=m-1+2m²-6m-2+m-5m²=-3m²-4m-3

原式=3(m^2-2m+1)+(m^3+8)-(m^3+3m^2-3m)=3m^2-6m+3+m^3+8-m^3-3m^2+3m=11-3m

|m|+|(2+m-m^2)除以(m+1)|=|m|+|(m^2-m-2)/(m+1)|=|m|+|(m-2)(m+1)/(m+1)|=|m|+|m-2|(1)当m>=2上式=m+m-2=2m-2(2

2[(m-1)m+m(m+1)]×[(m-1)m-m(m+1)]=2(m²-m+m²+m)×(m²-m-m²-m)=2×2m²×(-2m)=－8m&#

因为1+m^2-2m=(m-1)^2,因此原式=m+|(m-1)^2/(m-1)|=m+|m-1|,由于m≠1,因此当m>1时,原式=m+m-1=2m-1；当m

解m合适根号(3m+1)(2-m)=根号(3m+1)乘根号(2-m),即2-m＞0,且3m+1≥0即m＜2且3m+1≥0由|m-4|+|3m+1|+|m-2|=-（m-4）+3m+1-（m-2）=4-

2{（m-1)m+m(m+1)}{(m-1)m-m(m+1)}.=2（m²-m+m²+m）(m²-m-m²-m)=2*2m²*(-2m)=-8m