题目内容
已知a+b=0,a≠b,则化简
(a+1)+
(b+1)得( )
| b |
| a |
| a |
| b |
| A.2a | B.2b | C.+2 | D.-2 |
| b |
| a |
| a |
| b |
=b+
| b |
| a |
| a |
| b |
=(a+b)+(
| b |
| a |
| a |
| b |
又a+b=0,a≠b,可得:
| a |
| b |
| b |
| a |
∴
| b |
| a |
| a |
| b |
=(a+b)+(
| b |
| a |
| a |
| b |
=-2.
故选D.
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| b |
| a |
| a |
| b |
| A.2a | B.2b | C.+2 | D.-2 |
| b |
| a |
| a |
| b |
| b |
| a |
| a |
| b |
| b |
| a |
| a |
| b |
| a |
| b |
| b |
| a |
| b |
| a |
| a |
| b |
| b |
| a |
| a |
| b |