若a的倒数是-1/2√2,√b的相反数是0,c是-1的立方根,求c/(a-b)+a/(b-c)+b/(c-a)

Question

若a的倒数是-1/2√2,√b的相反数是0,c是-1的立方根,求c/(a-b)+a/(b-c)+b/(c-a)

Answer 1

因为 a 的倒数是-1/2√2 ,√b的相反数是0 ,c是-1的立方根
所以 a = -2√2 ,b = 0 ,c = -1
所以 c/(a - b) + a/(b - c) + b/(c - a)
= -1/(-2√2 - 0) + (-2√2)/(0 + 1) + 0/(-1 + 2√2 )
= 1/2√2 - 2√2 + 0
= √2/4 - 2√2
= -7√2/4