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(b-c)/(a²-ab-ac+bc)-(c-a)/(b²-bc-ab+ac)+(a-b)/(c²-ab-bc+ac)
=(b-c)/(a-b)(a-c)-(a-c)/(b-c)(a-b)+(a-b)/(a-c)(b-c)
=[(b-c)^2-(a-c)^2+(a-b)^2]/(a-b)(b-c)(a-c)
=[2b^2-2bc+2ac-2ab]/(a-b)(b-c)(a-c)
=2(a-b)(c-b)/(a-b)(b-c)(a-c)
=-2/(a-c)
1年前
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