We have : \(\dfrac{x^2-5x}{x-5}=5\)

\(\dfrac{x\left(x-5\right)}{x-5}=5\)

\(x=5\)

So the experiment of the above equation is \(S=\left\{5\right\}\)

We have : x2−5xx−5=5

x(x−5)x−5=5

x=5

So the experiment of the above equation is S={5}

Condition: \(x\ne5\)

We have: \(\dfrac{x^2-5x}{x-5}=5\Leftrightarrow\dfrac{x\left(x-5\right)}{x-5}=5\Leftrightarrow x=5\)

So the experiment of the above equation is \(S=\left\{\phi\right\}\)