设abc为已知有理数abc,c<b<0<a,化简...

设有理数abc在数轴上的对应点如图所示,化简│b-a│+│a+c│+│c-b│如图:c b 0 a
ruUS32MV99
│b-a│+│a+c│+│c-b│=a-b-c-a-c+b=-2c
那│b-a│+│a+c│-│c-b│呢?
│b-a│+│a+c│-│c-b│=a-b-c-a+c-b=-2b
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/etc/nginx/nginx.conf.已知有理数abc在数轴上的位置如图所示 —c—b—0—a,且|a|=|b| 化简|a|-|c-a|+|ac|-|-2b|
依题意:a>0,c-a<0,ac0则:|a|-|c-a|+|ac|-|-2b|=a+c-a-ac+2b=c-ac+2b
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a+a-c-ac-2+b这没过程,绝对值一脱就是这个结果了
看准正负号
记得变号就行了
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