Question

(1^4+1/4)*(3^4+1/4)*.........(19^4+1/4)/(2^4+1/4)*(4^4+1/4).....(20^4+1/4)等于多少

(1^4+1/4)*(3^4+1/4)*.........(19^4+1/4)/(2^4+1/4)*(4^4+1/4).....(20^4+1/4)这个怎么样计算啊,
怎么样计算呢,有什么简单的方法,这是在分式的通分 这一节出的题目.大家帮想下,谢了

Answer 1

x^4+1/4=x^4+x^2+1/4-x^2=(x^2+1/2)^2-x^2)
=(x^2+1/2+x)(x^2+1/2-x)=(x+1/2)^2+1/4)(x-1/2)^2+1/4)
即得x^4+1/4=(x+1/2)^2+1/4)(x-1/2)^2+1/4)
(1^4+1/4)*(3^4+1/4)*.........(19^4+1/4)/(2^4+1/4)*(4^4+1/4).....(20^4+1/4)
={((1/2)^2+1/4)((3/2)^2+1/4)((5/2)^2+1/4)((7/2)^2+1/4)...((37/2)^2+1/4)((39/2)^2+1/4)}/{((3/2)^2+1/4)((5/2)^2+1/4)((7/2)^2+1/4)((9/2)^2+1/4)...((39/2)^2+1/4)((41/2)^2+1/4)}
=((1/2)^2+1/4)/((41/2)^2+1/4)=2/(41^2+1)=1/841

Answer 2

1

Answer 3

可以用计算机找规律 我数学也不好