證明：
左邊=(b-c)/[(a-b)(a-c)]+(c-a)/[(b-c)(b-a)]+(a-b)/[(c-a)(c-b)]
=(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]/[(b-c)(a-c)(a-b)]
=(b^2-2bc+c^2+a^2-2ac+c^2+a^2-2ab+b^2)/[(b-c)(a-c)(a-b)]
=(2a^2+2b^2+2c^2-2ab-2bc-2ac)/[(b-c)(a-c)(a-b)]
=2(a^2+b^2+c^2-ab-bc-ac)/[(b-c)(a-c)(a-b)]
右邊=2/(a-b)+2/(b-c)+2/(c-a)
=2/(a-b)+2/(b-c)-2/(a-c)
=2[(a-c)(b-c)+(a-b)(a-c)-(a-b)(b-c)]/[(b-c)(a-c)(a-b)]
=2(ab-ac-bc+c^2+a^2-ac-ab+bc-ab+ac+b^2-bc)/[(b-c)(a-c)(a-b)]
=2(a^2+b^2+c^2-ac-ab-bc)/[(b-c)(a-c)(a-b)]
左＝右
下次出題不要弄得這么復雜,我都把題目算了好幾遍,才看清楚你的題目意思.不然的話,很少有人回答.