\(\frac{1+7y}{7x}=\frac{1+9y}{2x}\) \(\Leftrightarrow\frac{1+7y}{7}=\frac{1+9y}{2}\)

\(\Leftrightarrow\left(1+7y\right)2=7\left(1+9y\right)\)

\(\Leftrightarrow2+14y=7+63y\)

\(\Leftrightarrow63y-14y=2-7\)

\(\Leftrightarrow y=-\frac{5}{49}\)

Thay \(x=-\frac{5}{49}\) vào biểu thức ta có :

\(\frac{1+7.\frac{-5}{49}}{7.x}=\frac{1+9.\frac{-5}{49}}{2x}\)

\(\Leftrightarrow x=2\)

Vậy..

\(\frac{x}{5}=\frac{y}{9}\Rightarrow\hept{\begin{cases}x=5k\\y=9k\end{cases}}\)

\(\Rightarrow xy=45k^2\)  mà xy = 405

\(\Rightarrow45k^2=405\)

\(\Rightarrow k^2=9\)

\(\Rightarrow k=\pm3\)

thay k vào là được nhaa

Đặt \(\frac{x}{5}=\frac{y}{9}=k\Rightarrow x=5k,y=9\)\(k\)

\(\Rightarrow xy=45k^2\)mà \(xy=405\Rightarrow k^2=9\Rightarrow\hept{\begin{cases}k=9\\k=-9\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=45,y=81\\x=-45,y=-81\end{cases}}\)

                                  Vậy\(\hept{\begin{cases}x=45,y=81\\x=-45,y=-81\end{cases}}\)

Đặt : \(\frac{x}{5}=\frac{y}{9}=k\Rightarrow x=5k;y=9k\)

\(\Rightarrow x\cdot y=5k\cdot9k=45k^2=405\Rightarrow k^2=\frac{405}{45}=9\Rightarrow k=\pm3\)

TH1 : \(k=3\Rightarrow\hept{\begin{cases}\frac{x}{5}=3\Rightarrow x=3\cdot5=15\\\frac{y}{9}=3\Rightarrow y=3\cdot9=27\end{cases}}\)

TH2 : \(k=-3\Rightarrow\hept{\begin{cases}\frac{x}{5}=-3\Rightarrow x=-3\cdot5=-15\\\frac{y}{9}=-3\Rightarrow y=-3\cdot9=-27\end{cases}}\)

Vậy x = 15 thì y = 27; x = -15 thì y = -27

\(\text{Áp dụng tc dãy tỉ số bằng nhau ta có:}\frac{1+3y}{12}=\frac{1+7y}{4x}=\frac{2+10y}{12+4x}=\frac{1+5y}{6+2x}\)

\(+,y=-\frac{1}{5}\Rightarrow0=\frac{-2}{20x}\text{ vô lí}\)

\(\Rightarrow5y+1\ne0\Rightarrow6+2x=5x\Leftrightarrow x=2\Rightarrow\frac{1+3y}{12}=\frac{1+5y}{10}\Leftrightarrow5+15y=30y+6\Leftrightarrow y=\frac{-1}{15}\)