\(\dfrac{1+7y}{7x}=\dfrac{1+9y}{2x}\Rightarrow\dfrac{2+14y}{14x}=\dfrac{7+63y}{14x}\)

\(\Rightarrow2+14y=7+63y\Rightarrow49y=-5\Rightarrow y=\dfrac{-5}{49}\)

\(\Rightarrow\dfrac{1+5\left(\dfrac{-5}{49}\right)}{24}=\dfrac{1+7\left(\dfrac{-5}{49}\right)}{7x}\)

\(\Rightarrow\dfrac{1}{49}=\dfrac{2}{49x}\Rightarrow x=2\)

Vậy \(\left\{{}\begin{matrix}x=2\\y=\dfrac{-5}{49}\end{matrix}\right.\)

Theo bài ra: \(\dfrac{1+5y}{24}=\dfrac{1+7y}{7x}=\dfrac{1+9y}{2x}\left(1\right)\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\dfrac{1+5y}{24}=\dfrac{1+7y}{7x}=\dfrac{1+9y}{2x}=\dfrac{-2y}{24-7x}=\dfrac{-2y}{5x}\)

TH1: \(y=0\)

\(\Rightarrow\left(1\right)\Rightarrow\dfrac{1}{24}=\dfrac{1}{7x}=\dfrac{1}{2x}\) (vô lí)

\(\Rightarrow\) Loại

TH2: \(y\ne0\)

\(\Rightarrow\dfrac{-2y}{24-7x}=\dfrac{-2y}{5x}\)

\(\Rightarrow24-7x=5x\)

\(\Rightarrow12x=24\)

\(\Rightarrow x=2\)

Thay \(x=2\) vào \(\dfrac{1+5y}{24}=\dfrac{1+7y}{7x}\) , ta được:

\(\dfrac{1+5y}{24}=\dfrac{1+7y}{14}\)

\(\Rightarrow\left(1+5y\right)14=\left(1+7y\right)24\)

\(\Rightarrow14+70y=24+168y\)

\(\Rightarrow70y-168y=24-14\)

\(\Rightarrow-98y=10\)

\(\Rightarrow y=-\dfrac{5}{49}\)

Vậy \(x=2;y=-\dfrac{5}{49}\)

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

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

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

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

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

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..

áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{1+3y}{12}=\dfrac{1+5y}{5x}=\dfrac{1+7y}{4x}=\dfrac{1+7y-1-5y}{4x-5x}=\dfrac{2y}{-x}=\dfrac{1+5y-1-3y}{5x-12}=\dfrac{2y}{5x-12}\)

=>\(\dfrac{2y}{-x}=\dfrac{2y}{5x-12}\) với y=0 thay vào không thỏa mãn

nếu y khác 0

=>-x=5x-12

=>x=2. Thay x=2 vào trên ta được

\(\dfrac{1+3y}{12}=\dfrac{2y}{-2}=-y=>1+3y=-12y=>1=-15y=\dfrac{-1}{15}\)

Vậy x=2,y=\(\dfrac{-1}{15}\) thỏa mãn đề bài

1,7y

\(\dfrac{1+5y}{5x}=\dfrac{1+7y}{4x}\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\4\left(1+5y\right)=5\left(1+7y\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\15y=-1;y=-\dfrac{1}{15}\end{matrix}\right.\)

\(\dfrac{1+3y}{12}=\dfrac{1+5y}{x}\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x=\dfrac{12\left(1+5y\right)}{1+3y}=4.5.\left(\dfrac{3+15y}{5+15y}\right)=4.5.\left(\dfrac{3-1}{5-1}\right)=10\end{matrix}\right.\)\(\left(x;y\right)=\left(10;-\dfrac{1}{15}\right)\)