Each of the following expressions contains an instance of the variable \({\color{saddlebrown} h}\) or \({\color{purple} q}\) in their denominator. This instance of \({\color{saddlebrown} h}\) or \({\color{purple} q}\) is superfluous though. “Simplify” each expression to cancel the superfluous \({\color{saddlebrown} h}\) or \({\color{purple} q}\) and tidy the result at much as possible. You can check your answer by evaluating the initial expression and your result at some simple values of the variables (including \({\color{saddlebrown} h}\) or \({\color{purple} q}\)) to ensure they’re equal.
\(\displaystyle \frac{\bigl(2(3+h)-7\bigr)+1}{\color{saddlebrown} h} \)
\(\displaystyle \frac{(3+h)^2-(3)^2}{\color{saddlebrown} h} \)
\(\displaystyle \frac{\bigl(6(x+h)-5\bigr)-(6x-5)}{\color{saddlebrown} h} \)
\(\displaystyle \frac{\tfrac{1}{x+h}-\tfrac{1}{x}}{\color{saddlebrown} h} \)
\(\displaystyle \frac{(x+h)^2-(x-h)^2}{2{\color{saddlebrown} h}} \)
\(\displaystyle \frac{\sqrt{x+h}-\sqrt{x}}{\color{saddlebrown} h} \)
\(\displaystyle \frac{(x+h)^3-x^3}{\color{saddlebrown} h} \)
\(\displaystyle \frac{3(x+h)-(x+h)^2-\bigl(3x-x^2\bigr)}{\color{saddlebrown} h} \)
\(\displaystyle \frac{\tfrac{1}{1-(x+h)}-\tfrac{1}{1-x}}{\color{saddlebrown} h} \)
\(\displaystyle \frac{\sqrt{7+(x+h)}-\sqrt{7+x}}{\color{saddlebrown} h} \)
\(\displaystyle \frac{\tfrac{1}{(x+h)^2}-\tfrac{1}{x^2}}{\color{saddlebrown} h} \)
\(\displaystyle \frac{\sqrt{1-(x+h)^2}-\sqrt{1-x^2}}{\color{saddlebrown} h} \)
\(\displaystyle \frac{\tfrac{1}{\sqrt{x+h}}-\tfrac{1}{\sqrt{x}}}{\color{saddlebrown} h} \)
\(\displaystyle \frac{\tfrac{1}{qx}-\tfrac{1}{x}}{{\color{purple} q}x-x} \)
\(\displaystyle \frac{(qx)^2-x^2}{{\color{purple} q}x-x} \)
\(\displaystyle \frac{\sqrt{qx}-\sqrt{x}}{{\color{purple} q}x-x} \)