若1/a + 1/b + 1/c = 1/a+b+c,证明:1/a5 +1/b5 + 1/c5 = 1/a5+b5+c5

1/a+1/b=1/(a+b+c)-1/c所以a+b/ab = -(a+b) / c(a+b+c)若a+b = 0那么1/a^n+1/b^n+1/c^n=1/a^n+b^n+c^n显然成立否则ab = -c(a+b+c)也就是(b+c)(a+c) = 0所以b+c=0或a+c=0同样有1/a^n+1/b^n+1/c^n=1/a^n+b^n+c^nn=5有1/a5 +1/b5 + 1/c5 = 1/a5+b5+c5