已知实数-2<a<-1,0<b<1,且a的绝对值>b的绝对值,化简√(a+1)平方+√(b-1)平方-(a+b)绝对值
人气:278 ℃ 时间:2020-03-22 07:45:43
解答
√(a+1)平方+√(b-1)平方-(a+b)绝对值√(a+1)平方= |a+1| 因为-2<a<-1 |a+1|=-1-a√(b-1)平方=|b-1| 因为0<b<1 |b-1|=1-b-(a+b)绝对值 因为-2<a<-1,0<b<1,且a的绝对值>b的绝对值所以-(a+b)...
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