(1) 化简 (x-c)/(x-a)(x-b)+(b-c)/(a-b)(x-b)+(b-c)/(b-a)(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)(3) 证明,若a+b+c=0,则1/(b方+c方-a方)+1/(c方+a方-b方)+1/(a方+b方-c方)=0

(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=-2bc c²+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