Ta có: \(x+y+1=2\Rightarrow x+y=1\)
Lại có: \(3x=5y=4z\)
\(\Rightarrow\frac{3x}{60}=\frac{5y}{60}=\frac{4x}{60}\)
\(\Rightarrow\frac{x}{20}=\frac{y}{12}=\frac{z}{15}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{x}{20}=\frac{y}{12}=\frac{z}{15}=\) \(\frac{x+y}{20 +12}=\frac{1}{32}\)
Do \(\left[\begin{matrix}\frac{x}{20}=\frac{1}{32}\\\frac{y}{12}=\frac{1}{32}\\\frac{z}{15}=\frac{1}{32}\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x=\frac{5}{8}\\y=\frac{3}{8}\\z=\frac{15}{32}\end{matrix}\right.\)
Vậy \(\left[\begin{matrix}x=\frac{5}{8}\\y=\frac{3}{8}\\z=\frac{15}{32}\end{matrix}\right.\)
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