大师作文网计算(6aⁿ 2-9aⁿ 1
来源：学生作业帮助网 编辑：作业帮 时间：2024/09/21 15:38:23
a+a-2分之3-1=(a+b)-(2分之3+1)=2a-2分之5
正确答案应该是2/3a*n-1
(-2a)×(1/4a^3)=[(-2)×1/4]×(a×a^3)=-1/2a^4
很高兴收到您的求助：原式=2*（a-3)*(a+1)/(a-3)*(a-3)×1/4*(a+2)×(a-4)*(a+1)/(a-3)*(a+2)=1/2*(a-4)=2a-8分之1
=(1/(3a+1)^2-1/(3a+1))/(a^2/3a^2+a)=(3a/(3a+1)^2)/(a^2/3a^2+a)=(3a/(3a+1)^2)/(a/(3a+1))=3/(3a+1)
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1/a(a+2)=1/2(1/a-1/(a+2)).1/(a+18)(a+20)=1/2(1/(a+18)-1/(a+20))加在一起得原式=1/2(1/a-1/(a+2)+1/(a+2)-1/(a+
a²-9(a+b)²=a²-9(a²+2ab+b²)=a²-9a²-18ab-9b²=-8a²-18ab-9b
（1）原式=a+2a(a−2)•(a−2)2a+2=a−2a；（2）原式=2x(x+2)−x(x−2)(x−2)(x+2)•(x+2)(x−2)x=x(x+3)(x+2)(x−2)•(x+2)(x−2
a³*1/a²＝a³/a²＝a
(a+1)(a+2)(a-3)=（a²+2a+a+2）（a-3）=(a²+3a+2)(a-3)=a三次方-3a²+3a²-9a+2a-6=a三次方-7a-6互相
是a³/(a-1)-a²-a-1吧?==+已知：a³-1=(a-1)(a²+a+1)则：a³/(a-1)=(a³-1)/(a-1)+1/(a
(a+3a+5a+.+2003a)-(2a+4a+6a+...+2004a)=(a-2a)+(3a-4a)+(5a-6a)+.+(2003a-2004a)=-a-a-a-a...-a-a(共2004个
(3a^（n+1）+6a^（n+2）-9a^n)/3a^（n-1）=a²+2a³-3a再问：能不能给过程？再答：(3a^（n+1）+6a^（n+2）-9a^n)/3a^（n-1）=
:(3a^n+2+6a^n+1-9a^n)÷3a^n-1=3a^n+2÷3a^n-1+6a^n+1÷3a^n-1-9a^n÷3a^n-1=a³+2a²-3a再问：(3a^(n+2)
(a+1)(a+2)(a+3)(a-6)=[(a+1)(a-6)][(a+2)(a+3)]=(a²-5a-6)（a²+5a+6）=（a²）²-（5a+6）
a^2/(1+a)+1-a=a^2/(1+a)+(1-a)(1+a)/(1+a)=a^2/(1+a)+(1-a^2)/(1+a)=(a^2+1-a^2)/(1+a)=1/(1-a)再问：不对，答案是1
2a(a-3)^1-6a^2(3-a)-8a(a-3)=2a(a-3)+6a^2(a-3)-8a(a-3)=2a(a-3)（1+3a-4）=2a(a-3)(3a-3)=6a(a-3)(a-1)----