(a-b)^2+(b-c)^2+(c-a)^2=(b+c-2a)^2+(c+a-2b)^2+(a+b-2c)^2求bc/(a^2+1)+(ca+1)/(b^2+1)+(ab+2)/(c^2+1)的值

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(a-b)^2+(b-c)^2+(c-a)^2=(b+c-2a)^2+(c+a-2b)^2+(a+b-2c)^2求bc/(a^2+1)+(ca+1)/(b^2+1)+(ab+2)/(c^2+1)的值
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(a-b)^2+(b-c)^2+(c-a)^2=(b+c-2a)^2+(c+a-2b)^2+(a+b-2c)^2求bc/(a^2+1)+(ca+1)/(b^2+1)+(ab+2)/(c^2+1)的值
(a-b)^2+(b-c)^2+(c-a)^2=(b+c-2a)^2+(c+a-2b)^2+(a+b-2c)^2求bc/(a^2+1)+(ca+1)/(b^2+1)+(ab+2)/(c^2+1)的值

(a-b)^2+(b-c)^2+(c-a)^2=(b+c-2a)^2+(c+a-2b)^2+(a+b-2c)^2求bc/(a^2+1)+(ca+1)/(b^2+1)+(ab+2)/(c^2+1)的值
zheli

csac

恩恩 不会

怎么????