(a+1/a)(a^2+1/a^2)(a^4+1/a^4)(a^8+1/a^8)(a^16+1/a^16)(a^32+1

(a+1/a)(a^2+1/a^2)(a^4+1/a^4)(a^8+1/a^8)(a^16+1/a^16)(a^32+1/a^32)(a^2-1)
lanrenren 1年前 已收到4个回答 举报

觅水禾鱼 花朵

共回答了17个问题采纳率:82.4% 举报

原式 = [原式 *(a-1/a)]/(a-1/a) 注:方差公式
=[(a^2-1/a^2)(a^2+1/a^2)(a^4+1/a^4)(a^8+1/a^8)(a^16+1/a^16)(a^32+1/a^32)(a^2-1)]/(a-1/a)
=[(a^4-1/a^4)(a^4+1/a^4)(a^8+1/a^8)(a^16+1/a^16)(a^32+1/a^32)(a^2-1)]/(a-1/a)
=[(a^8-1/a^8)(a^8+1/a^8)(a^16+1/a^16)(a^32+1/a^32)(a^2-1)]/(a-1/a)
=[(a^16-1/a^16)(a^16+1/a^16)(a^32+1/a^32)(a^2-1)]/(a-1/a)
=[(a^32-1/a^32)(a^32+1/a^32)(a^2-1)]/(a-1/a)
=[(a^64-1/a^64)(a^2-1)]/(a-1/a)
分子分母同时乘以a/[a^2-1]
原式=(a^64-1/a^64)*a
=a^65-1/a^63

1年前

8

minamikitaoka 幼苗

共回答了8个问题 举报

乘以(a-1/a),再除以(a-1/a),利用平方差公式
=(a-1/a)(a+1/a)(a^2+1/a^2)(a^4+1/a^4)(a^8+1/a^8)(a^16+1/a^16)(a^32+1/a^32)(a^2-1)/(a-1/a)
=(a^2-1/a^2)(a^2+1/a^2)(a^4+1/a^4)(a^8+1/a^8)(a^16+1/a^16)(a^32+1/a^32)(a...

1年前

2

马元元 精英

共回答了21805个问题 举报

原式=(a-1/a)(a+1/a)(a^2+1/a^2)(a^4+1/a^4)(a^8+1/a^8)(a^16+1/a^16)(a^32+1/a^32)(a^2-1)/(a-1/a)
=(a^2-1/a^2)(a^2+1/a^2)(a^4+1/a^4)(a^8+1/a^8)(a^16+1/a^16)(a^32+1/a^32)(a^2-1)/[(a^2-1)/a]
反复用平方差
=(a^64-1/a^64)(a^2-1)/[(a^2-1)/a]
=a(a^64-1/a^64)
=a^65-1/a^63

1年前

1

逐欲天下 果实

共回答了3760个问题 举报

先乘(a-1/a),再除以(a-1/a)
原式=(a-1/a)(a+1/a)(a²+1/a²)...(a^32+1/a^32)(a²-1)÷(a-1/a)
=(a²-1/a²)(a²+1/a²).....(a^32+1/a^32)(a²-1)÷(a²-1)/a
=......
=(a^32-1/(a^32)(a^32+1/a^32)*a
=(a^64-1/a^64)*a
=a^65-1/a^63

1年前

1
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