Từ \(\frac{5x-1}{3}=\frac{7y-6}{5}\) Áp dụng TC DTSBN ta có :

\(\frac{5x-1}{3}=\frac{7y-6}{5}=\frac{\left(5x-1\right)+\left(7y-6\right)}{3+5}=\frac{5x+7y-7}{8}=\frac{5x+7y-7}{4x}\)

\(\Rightarrow4x=8\Rightarrow x=2\)

\(\Rightarrow\frac{5.2-1}{3}=\frac{7y-6}{5}\)

\(\Leftrightarrow\frac{7y-6}{5}=3\)

\(\Rightarrow y=3\)

Vậy \(x=2;y=3\)

\(\frac{5x-1}{3}=\frac{7y-6}{5}\Rightarrow5\left(5x-1\right)=3\left(7y-6\right)\Rightarrow25x-5=21y-18\)

\(\Rightarrow21y=25x+13\Rightarrow7y=\frac{25x+13}{3}\)

Xét : \(\frac{5x+7y-7}{4x}=\frac{5x+\frac{25x+13}{3}-7}{4x}=\frac{10x-2}{3x}\)

\(\Rightarrow3x\left(5x-1\right)=3\left(10x-2\right)\Rightarrow15x^2-33x+6=0\)

\(\Rightarrow3\left(x-2\right)\left(5x-1\right)=0\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{5}\end{cases}}\)

Với x=2 , ta có : y=3

Với x =\(\frac{1}{5}\), ta có : y= \(\frac{6}{7}\)

x=2

Có (5x-1+7y-6)/(3+5)=(5x+7y-7)/4x

       (5x+7y-7)/8=(5x+7y-7)/4x

    => 8=4x

    => x=2

Giải:

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

\(\dfrac{5x-1}{3}=\dfrac{7y-6}{5}=\dfrac{5x+7y-7}{8}=\dfrac{5x+7y-7}{4x}\)

+) Xét \(5x+7y-7=0\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{5x-1}{3}=0\\\dfrac{7y-6}{5}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}5x-1=0\\7y-6=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=\dfrac{6}{7}\end{matrix}\right.\)

+) Xét \(5x+7y-7\ne0\)

\(\Rightarrow4x=8\Rightarrow x=2\)

Thay \(x=2\) vào \(\dfrac{5x-1}{3}=\dfrac{7y-6}{5}\)

\(\Rightarrow3=\dfrac{7y-6}{5}\)

\(\Rightarrow7y=21\Rightarrow y=3\)

Vậy nếu \(5x+7y-7=0\) thì \(x=\dfrac{1}{5};y=\dfrac{6}{7}\)

nếu \(5x+7y-7\ne0\) thì x = 2, y = 3