\(\Rightarrow\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{2x-y}{10-2}=\dfrac{16}{8}=2\\ \Rightarrow\left\{{}\begin{matrix}x=10\\y=4\end{matrix}\right.\)

\(2x=5y=4z\)

\(\Leftrightarrow\frac{2x}{20}=\frac{5y}{20}=\frac{4z}{20}\)

\(\Leftrightarrow\frac{x}{10}=\frac{y}{4}=\frac{z}{5}\)

\(\Leftrightarrow\frac{x}{10}=\frac{y}{4}=\frac{5z}{25}\)

Áp dụng t,c dãy tỉ số bằng nhau ta có :

\(\frac{x}{10}=\frac{y}{4}=\frac{5x}{25}=\frac{x-y+5z}{10-4+25}=\frac{16}{31}=\frac{1}{2}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{x}{10}=\frac{1}{2}\\\frac{y}{4}=\frac{1}{2}\\\frac{5z}{25}=\frac{1}{2}\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=5\\y=2\\x=2,5\end{cases}}\)

\(\frac{x}{6}=\frac{y}{9}\)

Áp dụng tính chất của dãy tỉ số bằng nhau :

\(\Rightarrow\frac{x}{6}=\frac{y}{9}=\frac{x-y}{6-9}=\frac{30}{-3}=-10\)

\(\Rightarrow\frac{x}{6}=-10\Rightarrow x=-60\)

\(\frac{y}{9}=-10\Rightarrow y=-90\)

\(\frac{x}{y}=\frac{5}{4}\Rightarrow\frac{x}{4}=\frac{y}{5}\)

Áp dụng tính chất DTSBN :

\(\Rightarrow\frac{x}{4}=\frac{y}{5}=\frac{18}{9}=2\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{4}=2\Rightarrow x=8\\\frac{y}{5}=2\Rightarrow y=10\end{cases}}\)

\(e,7x=3y\Rightarrow\frac{x}{3}=\frac{y}{7}\)

ADTCDTSBN:

\(\Rightarrow\frac{x}{3}=\frac{y}{7}=\frac{x-y}{3-7}=\frac{16}{-4}=-4\)

\(\frac{x}{3}=-4\Rightarrow x=-12\)

\(\frac{y}{7}=-4\Rightarrow y=-28\)

\(2x=5y=10z\)

\(\Rightarrow\frac{2x}{10}=\frac{5y}{10}=\frac{10z}{10}\)

\(\Rightarrow\frac{x}{5}=\frac{y}{2}=\frac{z}{1}\)

áp dụng tính chất dãy tỉ số bằng nhau

\(\frac{x}{5}=\frac{y}{2}=\frac{z}{1}=\frac{x+y+z}{5+2+1}=\frac{16}{8}=2\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{5}=2\Rightarrow x=10\\\frac{y}{2}=2\Rightarrow y=4\\\frac{z}{1}=2\Rightarrow z=2\end{cases}}\)

ĐKXĐ: \(x\ne0\)

\(\dfrac{16+3y}{2x}=\dfrac{1+5y}{x}\Rightarrow\dfrac{16+3y}{2}=1+5y\)

\(\Rightarrow16+3y=2+10y\)

\(\Rightarrow y=2\)

\(\Rightarrow\dfrac{11}{x}=\dfrac{-1}{4}\Rightarrow x=-\dfrac{11}{4}\)