Question

已知实数a,b满足ab＜0,则下列不等式成立的是（    ）

A.|a+b|＞|a-b|                           B.|a+b|＜|a-b|

C.|a-b|＜||a|-|b||                       D.|a-b|＜|a|+|b|

Answer 1

思路解析：∵ab＜0,∴a,b异号，令a=2,b=-3.

则|a+b|=|2-3|=1,|a-b|=|2-(-3)|=5,

1＜5,∴|a+b|＜|a-b|.

答案：B