Với các số thực a,b,c thỏa mãn (a-b)(b-c)(c-a)=2022(a+b)(b+c)(a+c). Tính P=a/a+b+b/b+c+c/a+c

ĐK: $a+b\ne 0,b+c\ne 0,a+c\ne 0$

Đặt $\begin{cases} x=\dfrac{a-b}{a+b}\\y=\dfrac{b-c}{b+c}\\z=\dfrac{c-a}{c+a} \end{cases}$

$\begin{cases} x+1=\dfrac{a+b+a-b}{a+b}=\dfrac{2a}{a+b}\\y+1=\dfrac{b+c+b-c}{b+c}=\dfrac{2b}{b+c}\\z+1=\dfrac{c+a+c-a}{c+a}=\dfrac{2c}{c+a}\end{cases}\\\to (x+1)(y+1)(z+1)=\dfrac{8abc}{(a+b)(b+c)(a+c)}(1)\\ \begin{cases} 1-x=\dfrac{a+b-a+b}{a+b}=\dfrac{2b}{a+b}\\1-y=\dfrac{b+c-b+c}{b+c}=\dfrac{2c}{b+c}\\1-z=\dfrac{c+a-c+a}{c+a}=\dfrac{2a}{c+a} \end{cases}\\\to (1-x)(1-y)(1-z)=\dfrac{8abc}{(a+b)(b+c)(a+c)}(2)\\(1),(2)\to (x+1)(y+1)(z+1)=(1-x)(1-y)(1-z)\\\to (xy+x+y+1)(z+1)=(1-y-x+xy)(1-z)\\\to xyz +xz +yz +z +xy+x+y+1= 1-y-x+xy - z +yz+xz-xyz\\\to 2(xyz+x+y+z)=0\\\to xyz+x+y+z=0\\\to x+y+z=-xyz\\xyz=\dfrac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(a+c)}=\dfrac{2022(a+b)(b+c)(a+c)}{(a+b)(b+c)(a+c)}=2022\\\to -xyz=-2022\\\begin{cases} x+1=\dfrac{2a}{a+b}\\y+1=\dfrac{2b}{b+c}\\z+1=\dfrac{2c}{c+a} \end{cases}\to \begin{cases} \dfrac{x+1}{2}=\dfrac{a}{a+b}\\\dfrac{y+1}{2}=\dfrac{b}{b+c}\\\dfrac{z+1}{2}=\dfrac{c}{a+c} \end{cases}\\\to P=\dfrac{x+y+z+3}{2}=\dfrac{-2022 +3}{2}=\dfrac{-2019}{2}$