【初三数学】一道有规律的计算题》》

来源：百度知道 编辑：UC知道 时间：2024/06/08 12:35:36
计算：
(a+1/a)(a②+1/a②)(a④+1/a④)(a⑧+1/a⑧)(a拾陆+1/a拾陆)(a②-1)

其中的④之类的是前面的数的次方，“拾陆”也是。

计算出来附带过程就可以了，谢谢！

用平方差
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^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^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^2-1)/(a-1/a)
=(a^8-1/a^8)(a^8+1/a^8)(a^16+1/a^16)(a^2-1)/(a-1/a)
=(a^16-1/a^16)(a^16+1/a^16)(a^2-1)/(a-1/a)
=(a^32-1/a^32)*(a^2-1)/[(a^2-1)/a]
=a(a^32-1/a^32)
=a^33-1/a^31

(a+1/a)(a②+1/a②)(a④+1/a④)(a⑧+1/a⑧)(a拾陆+1/a拾陆)(a②-1)
=a（a-1/a)(a+1/a)(a②+1/a②)(a④+1/a④)(a⑧+1/a⑧)(a拾陆+1/a拾陆)
=a[(a^2-1/a^2)(a②+1/a②)(a④+1/a④)(a⑧+1/a⑧)(a拾陆+1/a拾陆)]
=a[(a^4-1/a^4)(a④+1/a④)(a⑧+1/a⑧)(a拾陆+1/a拾陆)]
=a[(a^8-1/a^8)(a^8+1/a^8)(a16+1/a^16)]
=a[a^32-1/a^32]
=a^33-1/a^31

统一一下表达：a的n次方写作a^n

乘一个（a-1/a）,再除一个(a-1/a),变成了
{(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^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^2-1)}/(a-1/a)
={(a^4-1/a^4)(a^4+1/a^4)(a^8+1/a