Question

已知a+b+c=0且a^2/(y-z)+b^2/(z-x)+c^2/(x-y)=0,求(ax^2+

Answer 1

问：已知a+b+c=0且a^2/(y-z)+b^2/(z-x)+c^2/(x-y)=0,求(ax^2+

答：你好，已知a+b+c=0且a^2/(y-z)+b^2/(z-x)+c^2/(x-y)=0,求(ax^2+解答：由c=-(a+b)可得2a²+bc=2a²-b(a+b)=2a²-ab-b²=(2a+b)(a-b)=(a-b)(a-c)同理2b²+ac=(b-c)(b-a); 2c²+ab=(c-a)(c-b)左式通分相加得:[a²(b-c)-b²(a-c)+c²(a-b)]/[(a-b)(a-c)(b-c)]将b=-(a+c)代入左式分子:[a²(b-c)-b²(a-c)+c²(a-b)]=(a-b)(a-c)(b-c)与分母相同∴a^2/(2a^2+bc)+b^2/(2b^2+ac)+c^2/(2c^2+ab)=1