Ta có: \(x=\dfrac{y}{8}=\dfrac{z}{27}\)
\(\Rightarrow\dfrac{3x}{3}=\dfrac{2y}{16}=\dfrac{7z}{189}\)
Áp dụng tính chất của dãy tỉ số bằng nhau:
\(\dfrac{3x}{3}=\dfrac{2y}{16}=\dfrac{7x}{189}=\dfrac{3x-2y+7z}{3-16+189}=\dfrac{5}{176}\)
Khi đó ta có: \(\left\{{}\begin{matrix}x=\dfrac{5}{176}\\\dfrac{y}{8}=\dfrac{5}{176}\\\dfrac{z}{27}=\dfrac{5}{176}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{176}\\y=\dfrac{5}{176}.8=\dfrac{5}{22}\\z=\dfrac{5}{176}.27=\dfrac{135}{176}\end{matrix}\right.\)
\(x=\dfrac{y}{8}=\dfrac{z}{27}\\ =>\dfrac{3x}{3}=\dfrac{2y}{16}=\dfrac{7z}{189}\)
mà `3x-2y+7z=5` nên áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\dfrac{3x}{3}=\dfrac{2y}{16}=\dfrac{7z}{189}=\dfrac{3x-2y+7z}{3-16+189}=\dfrac{5}{176}\\ =>x=\dfrac{5}{176}\\ =>y=\dfrac{5}{176}\cdot8=\dfrac{5}{22}\\ =>z=\dfrac{5}{176}\cdot27=\dfrac{135}{176}\)
Có \(x=\dfrac{y}{8}=\dfrac{z}{27}\\ \Rightarrow\dfrac{x}{1}=\dfrac{y}{8}=\dfrac{z}{27}\)
Áp dụng tính chất của DTSBN , ta có :
\(\dfrac{x}{1}=\dfrac{y}{8}=\dfrac{z}{27}=\dfrac{3x-2y+7z}{3\cdot1-2\cdot8+7\cdot27}=\dfrac{5}{176}\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{x}{1}=\dfrac{5}{176}\\\dfrac{y}{8}=\dfrac{5}{176}\\\dfrac{z}{27}=\dfrac{5}{176}\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{176}\\y=\dfrac{5}{22}\\z=\dfrac{135}{176}\end{matrix}\right.\)
