计算:b-c/(a-b)(a-c)-c-a/(b-c)(b-a)+a-b/(c-a)(c-b).(利用恒等式a±b/ab
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/24 11:16:11
计算:b-c/(a-b)(a-c)-c-a/(b-c)(b-a)+a-b/(c-a)(c-b).(利用恒等式a±b/ab=1/b±1/a简化运算)
=(b-c)/(a-b)(a-c)-(c-a)/(b-c)(b-a)+(a-b)/(c-a)(c-b)
=1/(a-b)-1/(a-c)-[1/(b-c)-1/(b-a)]+1/(c-a)-1/(c-b)
=1/(a-b)-1/(a-c)-1/(b-c)+1/(b-a)+1/(c-a)-1/(c-b)
=1/(a-b)+1/(b-a)+1/(c-a)-1/(a-c)-1/(b-c) -1/(c-b)
=1/(a-b)-1/(a-b)+1/(c-a)+1/(c-a)+1/(c-b) -1/(c-b)
=[1/(a-b)-1/(a-b)]+[1/(c-a)+1/(c-a)]+[1/(c-b) -1/(c-b)]
=0+2/(c-a)+0
=2/(c-a)
=1/(a-b)-1/(a-c)-[1/(b-c)-1/(b-a)]+1/(c-a)-1/(c-b)
=1/(a-b)-1/(a-c)-1/(b-c)+1/(b-a)+1/(c-a)-1/(c-b)
=1/(a-b)+1/(b-a)+1/(c-a)-1/(a-c)-1/(b-c) -1/(c-b)
=1/(a-b)-1/(a-b)+1/(c-a)+1/(c-a)+1/(c-b) -1/(c-b)
=[1/(a-b)-1/(a-b)]+[1/(c-a)+1/(c-a)]+[1/(c-b) -1/(c-b)]
=0+2/(c-a)+0
=2/(c-a)
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