Rút gọn phân thức sau: $\frac{bc+1}{a^{2}+2bc}+$ $\frac{ca+1}{b^{2}+2ac}+$ $\frac{ab+1}{c^{2}+2ab}$ (biết $\frac{1}{a}+$ $\frac{1}{b}+$ $\frac{1}{c}=0$)

Bổ sung đề : `a,b,c\ne 0`

`->abc\ne 0`

`->abc(1/a+1/b+1/c)=0`

`-> ab+ac+bc=0`

`-> ab=-ac-bc, ac=-ab-bc, bc=-ab-ac`

`a^2+2bc`

`=a^2+bc - ab-ac`

`=a(a-b)-c(a-b)`

`=(a-b)(a-c)`

`b^2+2ac`

`=b^2+ac - ab-bc`

`=b(b-a) + c (a-b)`

`= (a-b)(c-b)`

`c^2+2ab`

`=c^2+ab - ac-bc`

`= c (c-b) - a (c-b)`

`=(c-b)(c-a)`

`=-(a-c)(c-b)`

`(bc+1)/(a^2+2bc)+(ac+1)/(b^2+2ac)+(ab+1)/(c^2+2ab)`

`= (bc+1)/((a-b)(a-c)) + (ac+1)/((a-b)(c-b)) - (ab+1)/((a-c)(c-b))`

`= ((bc +1)(c-b) + (ac+1)(a-c) - (ab+1)(a-b))/((a-b)(a-c)(c-b))`

`= ((a-b)(a-c)(c-b))/((a-b)(a-c)(c-b))`

`=1`