20/02/22 13:31:35 InYZG21C.net
>>209
0 = 1/(a+x) - 2/(b+x) + 1/(c+x)
= {(2b-a-c)(x-b) + 2(bb-ac)}/{(a+x)(b+x)(c+x)}
= (2b-a-c)(x-b)/{(a+x)(b+x)(c+x)}, (← bb=ac)
2b-a-c≠0 より x=b
20/02/22 13:31:35 InYZG21C.net
>>209
0 = 1/(a+x) - 2/(b+x) + 1/(c+x)
= {(2b-a-c)(x-b) + 2(bb-ac)}/{(a+x)(b+x)(c+x)}
= (2b-a-c)(x-b)/{(a+x)(b+x)(c+x)}, (← bb=ac)
2b-a-c≠0 より x=b