Lời giải:

Từ $\frac{1+5y}{5x}=\frac{1+7y}{4x}$

$\Rightarrow \frac{1+5y}{5}=\frac{1+7y}{4}$

$\Rightarrow 4(1+5y)=5(1+7y)$

$\Rightarrow 4+20y=5+35y$

$\Rightarrow y=\frac{-1}{15}$

Thay vào điều kiện ban đầu:
$(1+3.\frac{-1}{15}):12=(1+5.\frac{-1}{15}):(5x)$

$\Rightarrow \frac{1}{15}=\frac{2}{15}:x$

$\Rightarrow x=2$

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

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

\(\Rightarrow\dfrac{1+5y}{5x}=\dfrac{1+7y}{4x}=\dfrac{1+5y-1+7x}{\left(5x-4x\right)}=\dfrac{-2y}{x}\)

\(\Rightarrow\dfrac{\left(1+5y\right)}{5}=-2y\)

Giải ra ta có: \(y=\dfrac{-1}{15}\)

\(\Leftrightarrow x=2\)

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

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

Khi đó \(\dfrac{1+5y}{5x}=\dfrac{-2y}{x}\)

\(\Rightarrow\left(1+5y\right)x=-10xy\)

\(\Rightarrow x+5xy=-10xy\)

\(\Rightarrow x=-10xy-5xy\)

\(\Rightarrow x=-15xy\)

\(\Rightarrow y=\dfrac{-1}{15}\)

và \(x=2\)

Vậy \(\left(x,y\right)=\left(2,\dfrac{-1}{15}\right)\).

Ta có : \(\dfrac{1+5y}{5x}\) = \(\dfrac{1+7y}{4x}\)

=> \(\dfrac{4\left(1+5y\right)}{20x}\) = \(\dfrac{5\left(1+7y\right)}{20x}\)

=> 4(1 + 5y) = 5(1 + 7y)

=> 4 + 20y = 5 + 35y

=> 4 - 5 = 35y - 20y

=> -1 = 15y

=> y = \(\dfrac{-1}{15}\)

Thay vào trên ta có : \(\dfrac{1+5y}{5x}\) = \(\dfrac{1}{15}\)

=> \(\dfrac{2}{3}\) : 5x = \(\dfrac{1}{15}\)

=> 5x = 10

=> x = 2

Vậy x = 2 và y = \(\dfrac{-1}{15}\)

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

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

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

\(\frac{3y+1}{12}\)=\(\frac{5y+2}{5x}\)=\(\frac{7y+3}{4x}\)=\(\frac{3y+1+7y+3}{12+4x}\)=\(\frac{4+10y}{12+4x}\)=\(\frac{2\left(2+5y\right)}{2\left(6+2x\right)}\)=\(\frac{2+5y}{6+2x}\)

\(\Leftrightarrow\)\(\frac{2+5y}{6+2x}\)=\(\frac{5y+2}{5x}\)

\(\Leftrightarrow\)6+2x=5x

\(\Leftrightarrow\)6=5x-2x=3x

\(\Leftrightarrow\)x=2

Thay x=2 vào \(\frac{5y+2}{5x}\) và \(\frac{7y+3}{4x}\) ta được:\(\frac{5y+2}{10}\)=\(\frac{7y+3}{8}\)

\(\Leftrightarrow\)8(5y+2)=10(7y+3)

\(\Leftrightarrow\)40y+16=70y+30

\(\Leftrightarrow\)40y-70y=30-16

\(\Leftrightarrow\)-30y=14

\(\Leftrightarrow\)y=\(\frac{-14}{30}\)

Vậy (x,y) là (2,\(\frac{-14}{30}\))

Từ \(\dfrac{1+5y}{5x}=\dfrac{1+7y}{4x}\Rightarrow\dfrac{4+20y}{20x}=\dfrac{5+35y}{20x}\)

\(\Rightarrow4+20y=5+35y\)

\(4-5=35y-20y\)

\(\Rightarrow15y=-1\)

\(\Rightarrow y=\dfrac{-1}{15}\)

Thay \(y=\dfrac{-1}{15}\) vào biểu thức ban đầu, ta được :

\(\dfrac{1+3\dfrac{-1}{15}}{12}=\dfrac{1+5\dfrac{-1}{15}}{5x}\)

\(\dfrac{\dfrac{4}{5}}{12}=\dfrac{\dfrac{2}{3}}{5x}\)

\(\Rightarrow12\dfrac{2}{3}=x\dfrac{4}{5}\)

\(x=12\dfrac{2}{3}:\dfrac{4}{5}=\dfrac{38}{3}\cdot\dfrac{5}{4}=\dfrac{95}{6}\)

Vậy ...

\(\dfrac{1+3y}{12}=\dfrac{1+5y}{5x}=\dfrac{5+15y}{60}=\dfrac{3+15y}{15x}=\dfrac{2}{60-15x}\)

\(\dfrac{1+3y}{12}=\dfrac{1+7y}{4x}=\dfrac{7+21y}{84}=\dfrac{3+21y}{12x}=\dfrac{4}{84-12x}\)

\(\Rightarrow\dfrac{2}{60-15x}=\dfrac{4}{84-12x}\Leftrightarrow168-24x=240-60x\)

\(\Leftrightarrow36x=72\Rightarrow x=2\)

\(\Rightarrow\dfrac{1+3y}{12}=\dfrac{2}{60-15.2}=\dfrac{2}{30}=\dfrac{1}{15}\)

\(\Leftrightarrow15+45y=12\Rightarrow45y=-3\Rightarrow y=\dfrac{-1}{15}\)

Vậy \(\left(x;y\right)=\left(2;\dfrac{-1}{15}\right)\)