Chủ đề này có 1 trả lời

[200dong]

Tìm x, y, z:

 

$\frac{x}{2y + 2z +1} = \frac{y}{2x + 2z +1} = \frac{z}{2x + 2y -2} = 2(x + y+z)$    

[DarkBlood]

 

Tìm x, y, z:

 

$\frac{x}{2y + 2z +1} = \frac{y}{2x + 2z +1} = \frac{z}{2x + 2y -2} = 2(x + y+z)$    

 

Trường hợp 1: $x+y+z=0$

Khi đó, ta có:

$$\frac{x}{2y + 2z +1} = \frac{y}{2x + 2z +1} = \frac{z}{2x + 2y -2} = 0$$

$$\Leftrightarrow x=y=z=0$$

 

Trường hợp 2: $x+y+z\neq 0$

Ta có:

$$\frac{x}{2y + 2z +1} = \frac{y}{2x + 2z +1} = \frac{z}{2x + 2y -2} = 2(x + y+z)$$

$$\Rightarrow \frac{x}{2y + 2z +1} = \frac{y}{2x + 2z +1} = \frac{z}{2x + 2y -2} =\frac{x+y+z}{4(x+y+z)}= 2(x + y+z)$$

$$\Rightarrow x+y+z= 8(x+y+z)^2$$

$$\Rightarrow x+y+z=\frac{1}{8}$$

Do đó: $$\frac{x}{2y + 2z +1} = \frac{y}{2x + 2z +1} = \frac{z}{2x + 2y -2}=\frac{1}{4}$$

$$\Leftrightarrow \left\{\begin{matrix} x=\frac{1}{2}y+\frac{1}{2}z+\frac{1}{4}\ \ \ \ \ \ \ \ \ \ (1)\\ y=\frac{1}{2}x+\frac{1}{2}z+\frac{1}{4}\ \ \ \ \ \ \ \ \ \ (2)\\ z=\frac{1}{2}x+\frac{1}{2}y-\frac{1}{2}\ \ \ \ \ \ \ \ \ \ (3) \end{matrix}\right.$$

Trừ $(1)$ với $(2),$ ta có:

$$\begin{aligned} &x-y=-\frac{1}{2}(x-y)\\ &\Leftrightarrow x-y=0 \\ &\Leftrightarrow x=y \end{aligned}$$

Cộng $(1)$ và $(2),$ ta có:

$$\begin{aligned} &x+y=\frac{1}{2}(x+y)+z+\frac{1}{2}\\\ &\Leftrightarrow x+y=z+\frac{1}{2}+z+\frac{1}{2}\\ &\Leftrightarrow 2x=2z+1 \end{aligned}$$

Do đó: $$\begin{aligned} &x+y+z=\frac{1}{8}\\\ &\Leftrightarrow 2x+z=\frac{1}{8}\\ &\Leftrightarrow 2z+1+z=\frac{1}{8}\\ &\Leftrightarrow z=-\frac{7}{24} \end{aligned}$$

Từ đó có $x=y=\frac{5}{24}$

 

Vậy $\boxed{\left (x\ ;\ y\ ;\ z \right )=\left (0\ ;\ 0\ ;\ 0 \right )\ ;\ \left ( \frac{5}{24}\ ;\ \frac{5}{24}\ ;\ -\frac{7}{24} \right )}$