b c

|c-b|+|a-c|-|b+c|,（c b c (-b+c)(b-c)过程 b>c且c [b/(a-b+c)]+[(2a+c)/(b-a-c)]-[(b-c)/(b-a-c)] 计算：a/[(a-b)(b-c)]+b/[(b-c)(b-a)]+c/[(c-a)(c-d)] b/a-b+c+2a+c/b-a-c-b-c/b-c-a 证明；（a+b)/(a-b)+(b+c)/(b-c)+(c+a)/(c-a)+[(a+b)(b+c)(c+b)/(a-b)(b-c)(c-a)]=0 |a+b|-|c| |a-c|-|b-a|+|a+c|c 化简|a-b|-2c-|c+b|+|3b| c ｜a+b｜+｜b-c｜-｜a+c｜+｜b+c｜怎么做 (a-b-c)(b+c-a)(c-a+b)= 请化简|a-b-c|+|b-c-a|+|c-a-b|. c语言(a>b)?(b>c?b:(a>c?c:a)):((a>c)?a:((b>c)?c:b))怎么看, 计算(c-a)/(a-b)(b-c)+(a-b)/(b-c)(c-a)+(b-c)/(c-a)(a-b) 行列式计算a b b b c a b b c c a b c c c a 化简:a+b/(a-c)(b-c)-b+c/(a-b)(c-a)+c+a/(c-b)(a-b) (a-b+c/a+b-c)-(a-2b+3c/b-c+a)+(b-2c/c-a-b)