计算a^2n+1-6a^2n+9a^2n-1/a^2-9除以a^n+1+4a^n+4a^n-1/a^2-4
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/30 05:42:06
x){nu
u%5R/g.y{ibPZDMR>&P赍N iH53 TRxTSJ
;Hij@56yv _ w
计算:(3a^n+2+6a^n+1-9^n)÷3a^n-1
分式 计算:a^(2n+1)-6a^(2n)+9a^(2n-1) / a^(n+1)-4a^n+3a^(n-1)
计算(3a^n+1-6a^n+9a^n-1)/(1/3a^n-2)
计算:(6a^n+2-9a^n+1+3a^n-1)÷3a^n-1
(3a^(n+1)+6a^(n+2)-9a^n)/3a^(n-1)是计算
计算:(-a^2)^n
(3a^n+1+6a^n+2-9a^n)/3a^n-1
(6a^n+2 +3a^n+1 -9a^n)/3a^n-1
计算(6a^(n+2)-8a^n)/(-1/2a^(n-1))
1.因式分解6(a-b)^2减8(a-b)^32.计算3a^(n+1)-6a^(n)+9a^(n-1)除以 三分之一a^(n-1
计算a^2n+1-6a^2n+9a^2n-1/a^2-9除以a^n+1+4a^n+4a^n-1/a^2-4
计算:(a^n+1)^2×(-a^3)^2n+1
计算 (3A^N+2*B-2A^N*B^N-1+3B^N)*5A^N*B^N+3(N为正整数,n>1)
设a,n∈N*证明a^2n-(-a)^n≥(a+1)×a^n
计算:(a(1)+a(2)+.+a(n-1))(a(2)+a(3)+.+a(n))-(a(2)+a(3)+.+a(n-1))(a(1)+a(2)+.a(n))
(4a^2n-6a^n+1+2a^n)/2a^n 因式分解
计算(3a的n+1次方-6a的n次方+9a的n-1次方)除以(3分之1a的n-2次方)
2道初二分式计算1.a^(n+1)-6a^n+9a^(n-1)2.a^(n+1)-4a^n+4a^(n-1)这两个怎么分解?化简