Question

已知a/|a|+|b|/b+c/|c|=1,求（|abc|/abc）一千九百九十九次方除以（bc/|ab|·ac/|bc

Answer 1

因为a/|a|+b/|b|+c/|c|=1

所以a,b,c中一定是两正一负

(1)a>0,b>0,c0,b0

|abc|/abc÷[(bc/|ab|)×(ac/|bc|×ab/|ac|)

=(-1)÷[(-c/a)×(-a/b)×(b/c)]

=(-1)÷1

=-1

(3)a0,c>0

|abc|/abc÷[(bc/|ab|)×(ac/|bc|×ab/|ac|)

=(-1)÷[(-c/a)×(a/b)×(-b/c)]

=(-1)÷1

=-1

综上|abc|/abc÷[(bc/|ab|)×(ac/|bc|×ab/|ac|)=-1