a-1分之a除以a²-1分之a²-a
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/17 23:11:03
(a-1分之a+1+a的平方-2a+1分之1)除以a-1分之a=[(a+1)/(a-1)+1/(a-1)²]×(a-1)/a=[(a+1)(a-1)/(a-1)²+1/(a-1)&
[1-1/(a+1)]/[(a^2-a)/(a+1)]=[a/(a+1)]/[a(a-1)/(a+1)]=1/(a-1)
解[1/(a+1)+1/(a-1)]÷2a/(a²-2a+1)=[(a-1)/(a-1)(a+1)+(a+1)/(a+1)(a-1)]×[(a-1)²/2a]=[(a-1+a+1)
(1/a-1)÷(a^2-1)/(a^2+a)=(1/a-1)÷[(a-1)(a+1)/a(a+1)]=(1-a)/a÷[(a-1)/a]=(1-a)/a*a/(a-1)=-(a-1)/a*a/(a-
(a-1/a)÷【(a²-2a+1)/a】=【(a²-1)/a】÷【(a-1)²/a】=【(a+1)(a-1)/a】*【a/(a-1)²】=(a+1)/(a-1
a+1/a+2=(√a)²+(1/√a)²+2*√a*(1/√a)=(√a+1/√a)²所以原式=√a+1/√a
a∧3-2a∧2+1/a∧4+a∧2-2a+1再问:符号不太清楚再答:你的是整体的平方吗a分之1-a平方是整体的吗是的话a立方-2a平方+1除以a四次方+a方-2a+1再问:不是,是÷a²+
[(a^2+2a分之a-2)-(a^2+4a+4分之a-1)]除以a+2分之a-4,其中a=-3=[(a-2)/a(a+2)-(a-1)/(a+2)²]÷(a-4)/(a+2)=[(a
(2+a-1分之1-a+1分之1)除以(a-1-a平方分之a)=[2+1/(a-1)-1/(a+1)]÷a[1-1/(1-a²)]=[2+1/(a-1)-1/(a+1)]÷a÷[(1-a
结果是b+a分之b-a再问:人
原式=(a-1/a)÷[(a-1)^2/a]=[(a^2-1)/a]÷[(a-1)^2/a]=[(a+1)(a-1)/a]/[(a-1)^2/a]=[(a+1)(a-1)]/(a-1)^2=(a+1)
=(a+1)/(a-1)-a/(a-1)²×a=(a+1)(a-1)/(a-1)²-a²/(a-1)²=(a²-1-a²)/(a-1)
1/8a8分之1除以a等于1/8*1/a=1/8a前面那位同学算的是8分之1乘以a
1/(a+1)+(a²-a)/(a²-2a-3)÷a/(a-3)=1/(a+1)+(a-1)/(a+1)=a/(a+1)再问:谢谢,我主要是问(a²-2a-3)怎样分解因
a分之1除以a=1∕a²a除以a分之1=a²8分之1除以a=1∕8aa分之1除以8=1∕8a
原式=[(a+1)/(a+1)(a-1)-(a-1)/(a+1)(a-1)-(a²-1)/(a+1)(a-1)]×(a+1)(a-1)=(a+1-a+1-a²+1)/(a+1)(a