计算:(b-c)/(a²-ab-ac+bc)-(c-a)/(b²-bc-ab+ac)+(a-b)/(c²-ab-bc+ab)
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计算:(b-c)/(a²-ab-ac+bc)-(c-a)/(b²-bc-ab+ac)+(a-b)/(c²-ab-bc+ab)
计算:(b-c)/(a²-ab-ac+bc)-(c-a)/(b²-bc-ab+ac)+(a-b)/(c²-ab-bc+ab)
计算:(b-c)/(a²-ab-ac+bc)-(c-a)/(b²-bc-ab+ac)+(a-b)/(c²-ab-bc+ab)
∵
a²-ab-ac+bc=(a-b)(a-c)
b²-bc-ab+ac=(b-a)(b-c)
c²-ac-bc+ab=(c-a)(c-b)
∴原式
=(b-c)/[(a-b)(a-c)]+(c-a)/[(b-a)(b-c)]+(a-b)/[(c-a)(c-b)]
=[(a-c)-(a-b)]/[(a-b)(a-c)]+[(b-a)-(b-c)]/[(b-a)(b-c)]+[(c-b)-(c-a)]/[(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/(c-a)+1/(b-c)+1/(a-b)+1/(c-a)+[1/(b-c)
=2[1/(a-b)]+2[1/(b-c)]+2[1/(c-a)]
=2[1/(a-b)+1/(b-c)+1/(c-a)]
化简到这一步已经可以,如要通分,则如下
=2{[(b-c)(c-a)+(a-b)(c-a)+(a-b)(b-c)]/[(a-b)(b-c)(c-a)]}
=2{[(bc-c²-ab+ac)+(ac-bc-a²+ab)+(ab-b²-ac+bc)]/[(a-b)(b-c)(c-a)]}
=2(ab+bc+ac-a²-b²-c²)/[(a-b)(b-c)(c-a)]
(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)