Firstly, we will factorise the quadratics by splitting middle terms

\(\displaystyle{x}^{{2}}+{2}{x}-{3}={x}^{{2}}+{3}{x}-{x}-{3}={x}{\left({x}+{3}\right)}-{1}{\left({x}+{3}\right)}={\left({x}-{1}\right)}{\left({x}+{3}\right)}\)

\(\displaystyle{x}^{{2}}+{x}-{20}={x}^{{2}}+{5}{x}-{4}{x}-{20}={x}{\left({x}+{5}\right)}-{4}{\left({x}+{5}\right)}={\left({x}-{4}\right)}{\left({x}+{5}\right)}\)

Now we will replace the quadratic by their factors and reciprocate the lower fraction.

\(\displaystyle\frac{{\frac{{{x}^{{2}}+{2}{x}-{3}}}{{{x}{2}+{x}-{20}}}}}{{\frac{{{x}-{1}}}{{{x}+{5}}}}}=\frac{{{\left({x}-{1}\right)}{\left({x}+{3}\right)}}}{{{\left({x}-{4}\right)}{\left({x}+{5}\right)}}}\cdot\frac{{{x}+{5}}}{{{x}-{1}}}=\frac{{{x}+{3}}}{{{x}-{4}}}\)

Ans:\(\displaystyle\frac{{{x}+{3}}}{{{x}-{4}}}\)

\(\displaystyle{x}^{{2}}+{2}{x}-{3}={x}^{{2}}+{3}{x}-{x}-{3}={x}{\left({x}+{3}\right)}-{1}{\left({x}+{3}\right)}={\left({x}-{1}\right)}{\left({x}+{3}\right)}\)

\(\displaystyle{x}^{{2}}+{x}-{20}={x}^{{2}}+{5}{x}-{4}{x}-{20}={x}{\left({x}+{5}\right)}-{4}{\left({x}+{5}\right)}={\left({x}-{4}\right)}{\left({x}+{5}\right)}\)

Now we will replace the quadratic by their factors and reciprocate the lower fraction.

\(\displaystyle\frac{{\frac{{{x}^{{2}}+{2}{x}-{3}}}{{{x}{2}+{x}-{20}}}}}{{\frac{{{x}-{1}}}{{{x}+{5}}}}}=\frac{{{\left({x}-{1}\right)}{\left({x}+{3}\right)}}}{{{\left({x}-{4}\right)}{\left({x}+{5}\right)}}}\cdot\frac{{{x}+{5}}}{{{x}-{1}}}=\frac{{{x}+{3}}}{{{x}-{4}}}\)

Ans:\(\displaystyle\frac{{{x}+{3}}}{{{x}-{4}}}\)