(1) 化简 (x-c)/(x-a)(x-b)+(b-c)/(a-b)(x-b)+(b-c)/(b-a)(x-a) (x-c)/(x-a)(x-b)+(b-c)/(a-b)(x-b)+(b-c)/(b-a)(x-a)= (x-c)/(x-a)(x-b)+ (b-c)/(a-b)•[1/(x-b)-1/(x-a)]=(x-c)/(x-a)(x-b)+ (b-c)/(a-b)•[（b-a）/(x-b) (x-a)]=(x-c)/(x-a)(x-b)- (b-c)/ /(x-b) (x-a)=(x-b) /(x-a)(x-b)=1//(x-a)(2) 化简(2a-b-c)/(a-b)(a-c)+(2b-c-a)/(b-c)(b-a)+(2c-a-b)/(c-b)(c-a) (2a-b-c)/(a-b)(a-c)+(2b-c-a)/(b-c)(b-a)+(2c-a-b)/(c-b)(c-a)=[(a-b)+(a-c)] /(a-b)(a-c)+[( b-c)+(b-a)]/(b-c)(b-a)+[(c-a)+(c-b)]/(c-b)(c-a)=1/(a-b)+1/(a-c)+1/(b-c)+1/(b-a)+1/(c-a)+1/(c-b)=1/(a-b)+1/(a-c)+1/(b-c)- 1/(a-b)-1/(a-c)-1/(b-c)=0(3) 证明,若a+b+c=0,则1/(b²+c²-a²)+1/(c²+a²-b²)+1/(a²+b²-c²)=0因为a+b+c=0,故：a=-(b+c),b=-(a+c),c=-(a+b)故：b²+c²-a²= b²+c²-[-(b+c)] ²=-2bcc²+a²-b²= c²+a²-[-( a+c)] ²=-2ac；a²+b²-c²＝a²+b²-[-( a+b)] ²=－2ab故：1/(b²+c²-a²)+1/(c²+a²-b²)+1/(a²+b²-c²)=1/(-2bc)+1/(-2ac)+1/(-2ab)=-(a+b+c)/(2abc)=0