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

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

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

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

-------- ( mk k chép lại đề nhé )

=> \(\dfrac{4+20y}{20x}=\dfrac{5+35y}{20x}\)

Với x khác 0 :

=> 4+20y=5+35y

4-5=35y-20y

-1=15y

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

Thay y=\(\dfrac{-1}{15}\) vào \(\dfrac{1+3y}{12}=\dfrac{1+5y}{5x}\) ta đc :

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

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

\(\dfrac{6}{5}=12:5x\)

5x=12:\(\dfrac{6}{5}\)

5x=10 => x=2

Vậy x=2 ; y =\(\dfrac{-1}{15}\)

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

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

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

Giải ra ta có: \(y=\dfrac{-1}{15}\)

\(\Leftrightarrow x=2\)

sao có x = 2

Ta có:

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

\(=\dfrac{2+10y}{2.\left(6+2x\right)}=\dfrac{2.\left(1+5y\right)}{2.\left(6+2x\right)}=\dfrac{1+5y}{6+2x}=\dfrac{1+5y}{5x}\)

- Xét \(1+5y=0\Rightarrow y=\dfrac{-1}{5}\Rightarrow1+5y=0\) ( loại )

- Xét \(1+5y\ne0\Rightarrow6+2x=5x\)

\(\Rightarrow5x-2x=6\)

\(\Rightarrow3x=6\)

\(\Rightarrow x=2\)

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

\(\Rightarrow\dfrac{1+3y}{12}=\dfrac{1+5y}{10}\)

\(\Rightarrow10.\left(1+3y\right)=12.\left(1+5y\right)\)

\(\Rightarrow10+30y=12+60y\)

\(\Rightarrow10-12=60y-30y\)

\(\Rightarrow-2=30y\)

\(\Rightarrow y=\dfrac{-1}{5}\)

Vậy \(x=2\) , \(y=\dfrac{-1}{5}\)

de ma

dễ mà k trả lời cũng như không dập tắt niềm tin của ng khác, thấy ghét, hứ