先化简,再求值:2m²-mn n²-4mn 4m²-搜答案-智慧作业帮

原式=9-m2+m2-6m-7=-6m+2=-6*1/2+2=-1

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

16m-6m-16m+2-8=-6m-6=0这个不难慢慢展开就做出来了,要动脑筋啊

m²-m分之m²-2m+1=(m-1)²/m(m-1)=(m-1)/m=(92-1)/92=91/92

很简单啊,合并相同的项：原式=-m^3+0m^2-2m=-m^3-2m带入m的值结果1-（-2）=3

4m−3-2m−6m2−6m+9=4m−3-2(m−3)(m−3)2=4m−3-2m−3=2m−3；当m=7时，原式=27−3=12．

劝自己做

原式=m+3+m−3(m+3)(m−3)•(m−3)22m=m−3m+3，当m=9时，原式=9−39+3=12．

M是2^(-x)的值域,结果为y>0,N是根号下x-1的值域,结果是y>=0两集合相交结果为{y|y>0}

(m-1)^2-3(2-5m+3m^2)=m^2-2m+1-6+15m-9m^2=-8m^2+13m-5=-(8m^2-13m+5)=-(m-1)(8m-5)将m=-1/3代入=-92/9

2分之3m-（2分之5m-1）+3(4-m)=2分之3m-2分之5m+1+12-3m=-4m+13=-4*3+13=1

原式＝m/(m+3)-6/(m+3)(m-3)/2/(m-3）=m/(m+3)-3/(m+3)=(m-3)/(m+3).

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

再求值m平方+m-2分之1+m除于m-2+m+2分之3

8m(m+n)²-2m(m-n)²=2m(4(m+n)²-(m-n)²)=2m(2m+2n+m-n)(2m+2n-m+n)=2m(3m+n)(3n+m)再把m=

原式={(m-1)²/[(m-1)(m+1)]}÷[(m-1)(m+1)/(m+1)-(m-1)/(m+1)]=[(m-1)/(m+1)]÷[m(m-1)/(m+1)]=[(m-1)/(m+

=3m^2-m-m^2+9+2(m^2-2m+1)=-5m+11将m=2代入,得：-10+11=1则原代数式值为1.

m(3m-1)-(m+3)(m-3)-2(m-1)^2=3m^2-m-(m^2-9)-2(m^2-2m+1)=3m^2-m-m^2+9-2m^2+4m-2=3m^2-m^2-2m^2-m+4m-2+9

原式=m+12m(m−1)•4m2(m+1)2-m+1−m+1(m+1)(m−1)=2m(m−1)(m+1)-2(m+1)(m−1)=2m−2(m+1)(m−1)=2m+1，当m=2时，原式=22+1

题目是[﹙m²-4m+4﹚/m²-1]÷﹙m-2/m-1+2/m-1﹚?还是m²-4m+﹙4/m²-1÷m﹚-2/m-1+2/m-1?或者是[m²-4