\(a,\dfrac{x-1}{x+5}=\dfrac{6}{7}\\ \Leftrightarrow\left(x-1\right).7=6\left(x+5\right)\\ \Rightarrow7x-7=6x+30\\ \Rightarrow7x-6x=7+30\\ \Rightarrow x=37\)
Vậy \(x=37\)
\(b,\dfrac{x^2}{6}=\dfrac{24}{25}\\ \Leftrightarrow x^2.25=24.6\\ \Rightarrow x^2.5^2=144\\ \Rightarrow\left(5x\right)^2=144\\ \Rightarrow\left(5x\right)^2=\left(\pm12\right)^2\\ \Rightarrow\left\{{}\begin{matrix}5x=12\\5x=-12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{12}{5}\\x=-\dfrac{12}{5}\end{matrix}\right.\)
Vậy \(x=\pm\dfrac{12}{5}\)