\(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{8}=\frac{y}{12}\) (nhân hai vế với \(\frac{1}{4}\) )
Tương tự : \(\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{15}\) (nhân hai vế với \(\frac{1}{3}\))
Suy ra \(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\)
Áp dụng tc dãy tỉ số bằng nhau : \(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y+z}{8+12+15}=\frac{10}{35}=\frac{2}{7}\)
Từ đó suy ra \(\begin{cases}x=\frac{16}{7}\\y=\frac{24}{7}\\z=\frac{30}{7}\end{cases}\)
\(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{8}=\frac{y}{12}\) ; \(\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{15}\)
Và x + y + z = 10
Áp dụng tính chất cua dãy tỉ số bằng nhau ta có:
\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\) \(=\frac{x+y+z}{8+12+15}=\frac{10}{35}=\frac{2}{7}\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{2}{7}.8\\y=\frac{2}{7}.12\\z=\frac{2}{7}.15\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{16}{7}\\y=\frac{24}{7}\\z=\frac{30}{7}\end{array}\right.\)
Vậy \(x=\frac{16}{7};y=\frac{24}{7};z=\frac{30}{7}\)
Ta có: \(\frac{x}{2}\) = \(\frac{y}{3}\) => \(\frac{x}{8}\) = \(\frac{y}{12}\)
\(\frac{y}{4}\) = \(\frac{z}{5}\) => \(\frac{y}{12}\) = \(\frac{z}{15}\)
=> \(\frac{x}{8}\) = \(\frac{y}{12}\) = \(\frac{z}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{x}{8}\) = \(\frac{y}{12}\) = \(\frac{z}{15}\) => \(\frac{x+y+z}{8+12+15}\) = \(\frac{10}{35}\) = \(\frac{2}{7}\)
=> \(\frac{x}{8}\) = \(\frac{2}{7}\) => x= \(\frac{2}{7}\) .8= \(\frac{16}{7}\)
\(\frac{y}{12}\) = \(\frac{2}{7}\) => y= \(\frac{2}{7}\) .12= \(\frac{192}{7}\)
\(\frac{z}{15}\) = \(\frac{2}{7}\) => z= \(\frac{2}{7}\) .15= \(\frac{2880}{7}\)
Vậy x = \(\frac{16}{7}\)
y = \(\frac{192}{7}\)
z = \(\frac{2880}{7}\)