標題:
[F.4]Maths Log
發問:
how to solve 3(5^2x-1)+5^2x+1=41?
最佳解答:
其他解答:
3(52??1)＋52?＋1＝41 ==> 3(52?)＋5(52?)＋5＝205 ?? (multiply each term by 5) ==> 8(52?)＝200 ==> 52?＝25＝52 ∴ 2x＝2 ==> x＝1|||||3 [ 5^(2x-1) ] + 5^(2x+1) = 41 or 3 [ 5^(2x) - 1 ] + 5^(2x) + 1 = 41 ?
[F.4]Maths Log
發問:
how to solve 3(5^2x-1)+5^2x+1=41?
最佳解答:
此文章來自奇摩知識+如有不便請留言告知
3 [5^(2x- 1)] + 5^(2x) + 1 = 41 3 [5^(2x - 1)] + 5^[1 + (2x - 1)] = 40 3 [5^(2x - 1)] + 5 [5^(2x - 1)] = 40 (3 + 5) [5^(2x - 1)] = 40 8 [5^(2x - 1)] = 40 5^(2x - 1) = 5 2x - 1 = 1 2x = 2 x = 1其他解答:
3(52??1)＋52?＋1＝41 ==> 3(52?)＋5(52?)＋5＝205 ?? (multiply each term by 5) ==> 8(52?)＝200 ==> 52?＝25＝52 ∴ 2x＝2 ==> x＝1|||||3 [ 5^(2x-1) ] + 5^(2x+1) = 41 or 3 [ 5^(2x) - 1 ] + 5^(2x) + 1 = 41 ?
文章標籤
全站熱搜
留言列表
