Ta có P=\(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\) =>abc.P=(a-b)ab +(b-c)bc +(c-a)ac
=> abc.P=(a-b)ab +bc(b-a+a-c) +(c-a)ac =(a-b)ab -(a-b)bc -bc(c-a) +(c-a)ac =(a-b)(ab-bc) +(c-a)(ac-bc) =(a-b).b.(a-c) -(a-c).c.(a-b) =(a-b)(c-a)(b-c) =>P= \(\dfrac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{abc}\)
Ta có Q= \(\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}\) =>(a-b)(c-a)(b-c) Q=a.(a-b)(c-a) +b(a-b)(b-c) +c(c-a)(b-c) = a(a-b)(c-a) +(-a-c)(a-b)(b-c) +c(c-a)(b-c) (b=-a-c)
= a(a-b)(c-a)-a(a-b)(b-c)-c(a-b)(b-c) +c(c-a)(b-c)
= a(a-b)(c-a-b+c) +c(b-c)(-a+b+c-a) =a(a-b)(2c-a-b) +c(b-c)(-2a+b+c) Vì a+b+c =0 =>a=-c-b ,b=-c-a ,c=-a-b =>(a-b)(c-a)(b-c)Q=a(a-b).3c +c(b-c).-3a =3ac(a-b) +3ac(-b+c) =3ac(a-2b+c)=3ac(-3b) =-9(abc) => Q=\(\dfrac{-9abc}{\left(a-b\right)\left(c-a\right)\left(b-c\right)}\)=\(\dfrac{9abc}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)=>P.Q= 9
