1-M 3分之3M-9除以M-3分之M的平方-6M 9,其中M=-3
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解题思路:利用分式的运算求解。解题过程:过程请见图片。最终答案:略
若m等于它的倒数即m=1/mm²=1m=±1(m²+m-6)/(m-2)÷(m+3)/(m²-3m+1)=[(m+3)(m-2)/(m-2)]*[(m²-3m+
分子:(m+1)(m+3)(2m-m^2)=m(m+1)(m+3)(2-m)=-m(m+1)(m+3)(m-2)分母:(m^3+m^2)(m-2)(m+3)=m^2(m+1)(m-2)(m+3)进行约
原式=m/(m+3)+6/(m+3)(m-3)*(m-3)/2=m/(m+3)+3/(m+3)=(m+3)/(m+3)=1
原式=2m3+3m-m3-5m+3m3-1=4m3-3m+3m-1=4m3-2m-1,当m=-1时,代入4m3-2m-1=-3.
已知m2次方+3m+1=0,m^2=-3m-1m3次方-8m的值=m(m^2-8)=m(-3m-1-8)=-3m(m+3)=-3m^2-9m=-3(-3m-1)-9m=9m+3-9m=3再问:=-3m
因为m^2+m-1=0把两边同时乘以m得到m^3+m^2-m=0再加上原等式m^2+m-1=0得到m^3+2m^2-1=0所以m3+2m2+2010=2011
=m/(m+3)-6/(m²-9)*(m-3)/2=m/(m+3)-3/(m+3)=(m-3)/(m+3)=-5/1=-5再问:题问错了,应该是。先化简,再求值:x+y分之x的平方-y的平方
(m-3除以3m的2次方-6m)除以(m+2-m-2分之5)=(m-3)/(3m^2-6m)÷[m+2-5/(m-2)]=(m-3)/3m(m-2)÷[(m+2)(m-2)-5]/(m-2)=(m-3
(1+n/m-m/m-n)=m(m-n)+n(m-n)-m]/[m(m-n)]=(m^2-n^2-m)/[m(m-n)](1-n/m-m/m+n)=m(m+n)-n(m+n)-m]/[m(m-n]=(
=m/(m+3)-6/(m²-9)*(m-3)/2=m/(m+3)-3/(m+3)=(m-3)/(m+3)=-5/1=-5
(M-3分之1)(M的平方+9分之1)(M+3分之1)=(M-3分之1)(M+3分之1)(M的平方+9分之1)=(M²-9分之1)(M²+9分之1)=M的4次方-81分之1很高兴为
∵m²+2m+1=0∴m=-1∴m³+2m²+3m=-2
3-m/2m-4除以(m+2-5/m-2)=(3-m)/2(m-2)÷((m+2)(m-2)-5)/(m-2)=(3-m)/2(m-2)÷(m²-9)/(m-2)=(3-m)/2(m-2)×
原式=m²(m-1)/m(m-1)÷(1+m)(1-m)/(m+1)=m/(1-m)在我回答的右上角点击【评价】,然后就可以选择【满意,问题已经完美解决】了.
[m/(m+3)-6/(m^2-9)]÷[2/(m-3)]=[m/(m+3)-6/(m+3)(m-3)]×[(m-3)/2]={[m(m-3)-6]/[(m+3)(m-3)]}×[(m-3)/2]=(
=(m-4)/(m+3)(m-3)×(m²+2m-3)/(m²-8m+16)×(m-3)=(m-4)/(m+3)×(m+3)(m-1)/(m-4)²=(m-1)/(m-4