[Toán 6] Tính chất của dãy tỉ số bằng nhau

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ngochienpc2000

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nguyenbahiep1

[laTEX]\frac{x}{y+z+1}= \frac{y}{x+z+1}= \frac{z}{x+y-2}= x+y+z \\ \\ \frac{x}{y+z+1}= \frac{y}{x+z+1}= \frac{z}{x+y-2}= \frac{x+y+z}{y+z+x+z+x+y+1+1-2} \\ \\ \Rightarrow \frac{x+y+z}{2(x+y+z)} =\frac{1}{2}= x+y+z \Rightarrow x+y+z = \frac{1}{2} \Rightarrow x+y = \frac{1}{2} - z \\ \\ \frac{x}{y+z+1}= \frac{y}{x+z+1}= \frac{x+y}{x+y+z+z+2} = \frac{1}{2} \\ \\ \Rightarrow \frac{\frac{1}{2} - z}{\frac{1}{2} + z+2} = \frac{1}{2} \Rightarrow z = - \frac{1}{2}[/laTEX]

tương tự tìm được y và x
 
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harrypham

Với $x+y+z=0 \implies x=y=z=0$.
Với $x+y+z \ne 0$.
$$\begin{array}{l} \frac{x}{y+z+1}= \frac{y}{x+z+1}= \frac{z}{x+y-2}= x+y+z \\ \frac{x}{y+z+1}= \frac{y}{x+z+1}= \frac{z}{x+y-2}= \frac{x+y+z}{y+z+x+z+x+y+1+1-2} \\ \implies \frac{x+y+z}{2(x+y+z)} =\frac{1}{2}= x+y+z \implies x+y+z = \frac{1}{2} \implies x+y = \frac{1}{2} - z \\ \frac{x}{y+z+1}= \frac{y}{x+z+1}= \frac{x+y}{x+y+z+z+2} = \frac{1}{2} \\ \implies \frac{\frac{1}{2} - z}{\frac{1}{2} + z+2} = \frac{1}{2} \implies z = - \frac{1}{2} \end{array}$$
 
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