已知实数a,b,c满足丨a-1丨+丨b+3丨+丨3c-1丨=0,求abc的125次方除以

因为绝对值大于等于零,所以三个绝对值均为零时结果才能为零
即丨a-1丨=0 丨b+3丨=0 丨3c-1丨=0
解得 a=1 b= -3 c = 1/3
故 a^9 = 1 b^3 = -27 c^= 1/9
abc = -3 a^9 * b^3 * c^= 1 *（-27）* 1/9 = -3
故(abc)^125 /(a^9 * b^3 * c^) = (-3)^125 /(-3)
= (-3)^124
= 3^124