((1^4+1/4)(3^4+1/4)(5^4+1/4).(19^4+1/4))/((2^4+1/4)(4^4+1/4)(6^4+1/4).(20^4+1/4))

来源:学生作业帮助网 编辑:作业帮 时间:2024/08/12 23:49:10
((1^4+1/4)(3^4+1/4)(5^4+1/4).(19^4+1/4))/((2^4+1/4)(4^4+1/4)(6^4+1/4).(20^4+1/4))
xRN@ܘδ[~~i$1"B!qMX&$ l?̴it;.ιsmk9W~- 6$T$D%K.rs6v"fpSN,%1MGԲ"H^a_-Y76b?ydstNFY|$N#5D 4˲&  HԸɀ8HhjR!'q:Ȳ. IQ!q |Jyq-:V^j2N\M3ņ#0 K6izԫanXiZYۯ.EYvLl0]Il<^_o5š&*EN

((1^4+1/4)(3^4+1/4)(5^4+1/4).(19^4+1/4))/((2^4+1/4)(4^4+1/4)(6^4+1/4).(20^4+1/4))
((1^4+1/4)(3^4+1/4)(5^4+1/4).(19^4+1/4))/((2^4+1/4)(4^4+1/4)(6^4+1/4).(20^4+1/4))

((1^4+1/4)(3^4+1/4)(5^4+1/4).(19^4+1/4))/((2^4+1/4)(4^4+1/4)(6^4+1/4).(20^4+1/4))
a^4+1/4=(a²+1/2)²-a²=(a²+a+1/2)(a²-a+1/2)=[a(a+1)+1/2][a(a-1)+1/2]
所以
[(1^4+1/4)(3^4+1/4)(5^4+1/4)……(19^4+1/4)]/[(2^4+1/4)(4^4+1/4)(6^4+1/4)……(20^4+1/4)]
=[(1*2+1/2)*(1*0+1/2)]·[(3*4+1/2)*(3*2+1/2)]·…·[(19*20+1/2)*(19*18+1/2)]/·[(2*3+1/2)*(2*1+1/2)]·[(4*5+1/2)*(4*3+1/2)]·…·[(20*21+1/2)*(20*19+1/2)]=(1*0+1/2)/(20*21+1/2)=1/841
无意中看到你的提问 就去百度了下 一不小心儿 还给找到了 就复制来了 具体对不对 你自己个儿 验证去吧!