131. \(\dfrac{\sqrt x}{\sqrt x-6}-\dfrac{3}{\sqrt x+6}+\dfrac{x}{36-x}\)

Положим \(t=\sqrt x\). Тогда \(x=t^2\), а \(36-x=36-t^2=-(t-6)(t+6)\).

\[\frac{t}{t-6}-\frac{3}{t+6}+\frac{t^2}{36-t^2}\]

\[=\frac{t(t+6)-3(t-6)-t^2}{(t-6)(t+6)}.\]

Раскрываем скобки:

\[\frac{t^2+6t-3t+18-t^2}{(t-6)(t+6)}=\frac{3t+18}{(t-6)(t+6)}.\]

\[\frac{3(t+6)}{(t-6)(t+6)}=\frac{3}{t-6}.\]

Ответ: \(\dfrac{3}{\sqrt x-6}\), где \(x\ge 0\), \(x\ne36\).