已知a+b+c=0,试求a^2/(2a^2+bc)+b^2/(2b^2+ac)+c^2/(2c^2+ab)的值提示:由已知a＝－(b+c)可得2a^2+bc＝a^2－a(b+c)+bc＝(a－b)(a－c)

Question

已知a+b+c=0,试求a^2/(2a^2+bc)+b^2/(2b^2+ac)+c^2/(2c^2+ab)的值
提示:由已知a＝－(b+c)可得2a^2+bc＝a^2－a(b+c)+bc＝(a－b)(a－c)

Answer 1

因为a+b+c=0,则b=-a-c,bc=-ac-c2
所以2a2+bc=2a2-ac-c2=(2a+c)(a-c)=(a-b)(a-c)
故:[a2/(2a2+bc)]+[b2/(2b2+ac)]+[C2/(2c2+ab)]
=[a2/(a-b)(a-c)]+[b2/(b-c)(b-a)]+[C2/(c-a)(c-b)]
=(a-b)*[a2/(a-c)-b2/(b-c)]+[C2/(c-a)(c-b)]
=(ac+c2)/(c-a)(c-b)+[C2/(c-a)(c-b)]
=(ac+2c2)/((c-a)(c-b))
=(ac+2c2)/(ac+2c2)
=1