Đặt \(\dfrac{1}{a}=\dfrac{1}{x+y},\dfrac{1}{b}=\dfrac{1}{y+z},\dfrac{1}{c}=\dfrac{1}{z+x}\)

Đề trở thành: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\), tính \(P=\dfrac{bc}{a^2}+\dfrac{ac}{b^2}+\dfrac{ab}{c^2}\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\) Tương đương \(ab+bc=-ac\)

\(P=\dfrac{b^3c^3+a^3c^3+a^3b^3}{a^2b^2c^2}=\dfrac{\left(ab+bc\right)\left(a^2b^2-ab^2c+b^2c^2\right)+a^3c^3}{a^2b^2c^2}=\dfrac{-ac\left(a^2b^2-ab^2c+b^2c^2\right)+a^3c^3}{a^2b^2c^2}\)

\(=\dfrac{a^2c^2-a^2b^2+ab^2c-b^2c^2}{ab^2c}=\dfrac{ac}{b^2}-\dfrac{a}{c}+1-\dfrac{c}{a}\)\(=ac\left(\dfrac{1}{a^2}+\dfrac{2}{ac}+\dfrac{1}{c^2}\right)-\dfrac{a}{c}+1-\dfrac{c}{a}\) (do \(\dfrac{1}{b}=-\dfrac{1}{a}-\dfrac{1}{c}\) tương đương \(\dfrac{1}{b^2}=\dfrac{1}{a^2}+\dfrac{2}{ac}+\dfrac{1}{c^2}\)) 

\(=3\)

Vậy P=3

\(\left(x+y+z\right).\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}\right)=\dfrac{2017}{672}\)

\(\Rightarrow\left(\dfrac{x+y+z}{x+y}+\dfrac{x+y+z}{y+z}+\dfrac{x+y+z}{x+z}\right)=\dfrac{2017}{672}\)

\(\Rightarrow1+\dfrac{z}{x+y}+1+\dfrac{x}{y+z}+1+\dfrac{y}{x+z}=\dfrac{2017}{672}\)

\(\Rightarrow3+\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}=\dfrac{2017}{672}\)

\(\Rightarrow\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}=\dfrac{2017}{672}-3=\dfrac{2017}{672}-\dfrac{2016}{672}=\dfrac{1}{672}\)

\(\Rightarrow C=\dfrac{1}{672}\)

Lời giải:
Nếu $x+y+z=0$ thì:

$\frac{x+y-z}{z}=\frac{-z-z}{z}=-2$

$\frac{y+z-x}{x}=\frac{-x-x}{x}=-2$

$\frac{z+x-y}{y}=\frac{-y-y}{y}=-2$ 

(thỏa mãn đkđb)

Khi đó:

$P=(1+\frac{x}{y})(1+\frac{y}{z})(1+\frac{z}{x})=\frac{(x+y)(y+z)(z+x)}{xyz}$

$=\frac{(-z)(-x)(-y)}{xyz}=\frac{-xyz}{xyz}=-1$

Nếu $x+y+z\neq 0$

Áp dụng TCDTSBN:

$\frac{x+y-z}{z}=\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z+y+z-x+z+x-y}{z+x+y}=\frac{x+y+z}{x+y+z}=1$

$\Rightarrow x+y=2z; y+z=2x, z+x=2y$. Khi đó:

$P=\frac{(x+y)(y+z)(z+x)}{xyz}=\frac{2z.2x.2y}{xyz}=8$

Xét : 2017.2017 = (x+y+z).(1/x+y + 1/x+z + 1/y+z)

= x/y+z + y/x+z + z/x+y + 1 + 1 + 1

= x/y+z + y/x+z + z/x+y + 3

=> A = x/y+z + y/x+z + z/x+y = 2017^2 - 3 = 4068286

Tk mk nha

Ta có :(x+y+z)(1/x+y  +  1/y+z  +  1/x+z) =2017²

=>x/x+y  +y/x+y  +z/x+y  +  x/y+z +  y/y+z +  z/y+z  +x/x+z  +  y/x+z  +  z/x+z=2017²

=>(x/x+y  +  y/x+y)+(y/y+z  +  z/y+z)+(x/x+z  +  z/x+z)+(x/y+z  +  y/x+z  +  z/x+y)    =4068289

=>1+1+1+A=4068289

=>A=4068286