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:

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

\(\frac{2016.x}{xy+2016x+2016}+\frac{y}{yz+y+2016}+\frac{z}{xz+z+1}\)= \(\frac{2016x}{xy+2016x+1}+\frac{xy}{xyz+xy+2016x}+\frac{xyz}{xxyz+xyz+xy}\)     = \(\frac{2016x}{xy+2016x+xyz}+\frac{xy}{xyz+xy+2016x}+\frac{xyz}{2016x+xyz+xy}\)

=\(\frac{2016x+xy+xyz}{2016x+xy+xyz}=1\)

thay xyz=2017, ta có:

\(D=\frac{xyzx}{xy+xyzx+xyz}+\frac{y}{yz+y+xzy}+\frac{z}{xz+z+1}\)

\(D=\frac{xz}{1+xz+z}+\frac{1}{x+1+xz}+\frac{z}{xz+x+1}=1\)

\(\text{Bài làm }\)

\(\text{ Gọi xyz = 2017}\)

\(\text{Ta có:}\) \(D=\frac{xyzx}{xy+xyzx+xyz}+\frac{y}{yz+y+xzy}+\frac{z}{xz+z+1}\)

           \(D=\frac{xz}{1+xz+z}+\frac{1}{x+1+xz}+\frac{z}{xz+x+1}=1\)

\(\text{# Chúc bạn học tốt #}\)

Ko sai bạn ey

{ x + y + z = 1 (1)

{ x² + y² + z² = 1 (2)

{ x³ + y³ + z³ = 1 (3)

(x + y + z)² = x² + y² + z² + 2(xy + yz + zx) 

⇒ 2(xy + yz + zx) = (x + y + z)² - (x² + y² + z²) = 1² - 1 = 0 ⇒ xy + yz + zx = 0

(x + y + z)³ = x³ + y³ + z³ + 3(x + y)(y + z)(z + x) 

⇒ 3(x + y)(y + z)(z + x) = (x + y + z)³ - (x³ + y³ + z³) = 1³ - 1 = 0

⇒ x + y = 0 hoặc y + z = 0 hoặc z + x = 0

@ Nếu  x + y = 0 ⇔ x = - y thay vào (1) ⇒ z = 1 , thay vào (2) ⇒ 2x² + 1 = 1 ⇒ x = 0; y = 0

⇒ S = 1

Tương tự cho trường hợp y + z = 0 và z + x = 0

ghse5uye5bvs

Đặt \(k=\frac{y+z-x}{7}=\frac{z+x-y}{11}=\frac{x+y-z}{5}=\frac{xyz}{3}\)

Áp dụng dãy tỉ số bằng nhau:

\(k=\frac{y+z-x}{7}=\frac{z+x-y}{11}=\frac{y+z-x+z+x-y}{7+11}=\frac{2z}{18}=\frac{z}{9}\)

=> z=9k

Tương tự:

\(k=\frac{x+y-z}{5}=\frac{z+x-y}{11}=\frac{2x}{16}=\frac{x}{8}\)

=> x=8k

\(k=\frac{x+y-z}{5}=\frac{y+z-x}{7}=\frac{2y}{12}=\frac{y}{6}\)

=> y=6k

Ta có: \(\frac{xyz}{3}=k\Rightarrow\frac{6k.9k.8k}{3}=k\Leftrightarrow144k^3-k=0\Leftrightarrow k\left(144k^2-1\right)=0\)

+) TH1: k=0 ta có: x=y=z=0

+) Th2: \(144k^2-1=0\Leftrightarrow k^2=\frac{1}{144}=\frac{1}{12^2}\Leftrightarrow k=\pm\frac{1}{12}\)

Với \(k=\frac{1}{12}\).

Ta có: \(z=9k=\frac{9}{12}=\frac{3}{4};x=8k=\frac{8}{12}=\frac{2}{3};y=6k=\frac{6}{12}=\frac{1}{2}\)

Với k=-1/12 Em tự tính nhé