题目内容
a |
(a-b)(a-c) |
b |
(b-c)(b-a) |
c |
(c-a)(c-b) |
A.a | B.b | C.1 | D.0 |
原式=
-
+
=
=
=
=0.
故选D.
a |
(a-b)(a-c) |
b |
(a-b)(b-c) |
c |
(a-c)(b-c) |
=
a(b-c)-b(a-c)+c(a-b) |
(a-b)(b-c)(a-c) |
=
ab-ac-ab+bc+ac-bc |
(a-b)(b-c)(a-c) |
=
0 |
(a-b)(b-c)(a-c) |
=0.
故选D.
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