\(\Leftrightarrow7\cdot\left(x-1\right)=6\cdot\left(x+5\right)\)
\(\Leftrightarrow7x-7-6x-30=0\)
\(\Leftrightarrow x-37=0\)
\(\Leftrightarrow x=37\left(N\right)\)
Ta có: \(\dfrac{x-1}{x+5}=\dfrac{6}{7}\)
\(\Leftrightarrow7x-7-6x-30=0\)
\(\Leftrightarrow x=37\)