Theo bài ra ta cs
\(2x=3y\Rightarrow\frac{x}{3}=\frac{y}{2}\)
\(4y=5z\Rightarrow\frac{y}{5}=\frac{z}{4}\)
T lại cs
\(\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{x}{15}=\frac{y}{10}\left(1\right)\)
\(\frac{y}{5}=\frac{z}{4}\Rightarrow\frac{x}{10}=\frac{z}{8}\left(2\right)\)
Từ (1);(2) \(\Rightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{8}\)
ADTC dãy tỉ số bằng nhau ta cs
\(\frac{x}{15}=\frac{y}{10}=\frac{z}{8}=\frac{2x+3y-4z}{2.15+3.10-4.8}=\frac{56}{28}=2\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{15}=2\\\frac{y}{10}=2\\\frac{z}{8}=2\end{cases}\Rightarrow\hept{\begin{cases}x=30\\y=20\\z=16\end{cases}}}\)
\(2x=3y;4y=5z\) => \(8x=12y;12y=15z\)
=> \(\frac{8x}{120}=\frac{12y}{120}=\frac{15z}{120}\)=> \(\frac{x}{15}=\frac{y}{10}=\frac{z}{8}\)
=> \(\frac{2x}{30}=\frac{3y}{30}=\frac{4z}{32}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{2x}{30}=\frac{3y}{30}=\frac{4z}{32}=\frac{2x+3y-4z}{30+30-32}=\frac{56}{28}\)
=> \(\frac{2x}{30}=2=>2x=60=>x=30\)
\(\frac{3y}{30}=2=>3y=60=>y=20\)
\(\frac{4z}{32}=2=>4z=64=>z=16\)
2x = 3y => \(\frac{x}{3}=\frac{y}{2}\) =>\(\frac{x}{15}=\frac{y}{10}\)
4y = 5z => \(\frac{y}{5}=\frac{z}{4}\) => \(\frac{y}{10}=\frac{z}{8}\)
=> x/15 = y/10 = z/8
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{15}=\frac{y}{10}=\frac{z}{8}\)=> \(\frac{2x}{30}=\frac{3y}{30}=\frac{4z}{32}=\frac{2x+3y-4z}{30+30-32}=\frac{56}{28}=2\)
=> \(\hept{\begin{cases}\frac{x}{15}=2\\\frac{y}{10}=2\\\frac{z}{8}=2\end{cases}}\) => \(\hept{\begin{cases}x=30\\y=20\\z=16\end{cases}}\)
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