Question

a>0,b>0,c>0 解方程 (x-a)/(b+c)+(x-b)/(c+a)+(x-c)/(a+b)=3x/(a+b+c) 急

Answer 1

先分解左边式子(x-a)/(b+c)+(x-b)/(c+a)+(x-c)/(a+b)=x(1/(b+c)+1/(c+a)+1/(a+b))-(a/(b+c)+b/(c+a)+c/(a+b))=x(1/(b+c)+1/(c+a)+1/(a+b))-((a+b+c)/(b+c)+(a+b+c)/(c+a)+(a+b+c)/(a+b)-3)=x(1/(b+c)+1/(c+a)+1/(a+b))-(a+b+c)(1/(b+c)+1/(c+a)+1/(a+b)-3/(a+b+c))
原式==>
x(1/(b+c)+1/(c+a)+1/(a+b))-(a+b+c)(1/(b+c)+1/(c+a)+1/(a+b)-3/(a+b+c))=3x/(a+b+c)
==>
x(1/(b+c)+1/(c+a)+1/(a+b))-3x/(a+b+c)= (a+b+c)(1/(b+c)+1/(c+a)+1/(a+b)-3/(a+b+c))
==>
x(1/(b+c)+1/(c+a)+1/(a+b)-3/(a+b+c))=(a+b+c)(1/(b+c)+1/(c+a)+1/(a+b)-3/(a+b+c))
得结果 x=a+b+c