第866题:建立方程求解
a,b,ca,b,ca,b,c 都是已知数,且
a+b+c=3a+b+c=3a+b+c=3
a(1b+1c)a(\dfrac{1}{b}+\dfrac{1}{c}) a(b1+c1)+b(1c+1a)+ b(\dfrac{1}{c}+\dfrac{1}{a}) +b(c1+a1)+c(1a+1b)+c(\dfrac{1}{a}+\dfrac{1}{b})+c(a1+b1)=3 =3=3
那么
1a+1b+1c \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}a1+b1+c1 =?=?=?
本题有提示.