Verified Answer :

\(\frac{(x+1)^2}{x-5}\cdot\frac{x-3}{x+1}=\frac{x+1}{x-5}\cdot\frac{x-3}{1}=\frac{(x+1)(x-3)}{x-5}=\frac{x^2-3x+x-3}{x-5}=\frac{x^2-2x-3}{x-5}\\\\D:x eq5\ \wedge\ x eq-1\)


\(\frac{3x+3}{2}:\frac{4}{9x+9}=\frac{3x+3}{2}\cdot\frac{9x+9}{4}=\frac{27x^2+27x+27x+27}{8}=\frac{27x^2+54x+27}{8}\\\\D:x eq1\)


\(\frac{4a}{3a-3}\cdot\frac{3(a-1)}{16a}=\frac{1}{3a-3}\cdot\frac{3a-3}{4a}=\frac{1}{4a}\\\\D:a eq1\ \wedge\ a eq0\)


\(\frac{x+2}{x-1}:\frac{5x+10}{x-2}=\frac{x+2}{x-1}\cdot\frac{x-2}{5x+10}=\frac{x+2}{x-1}\cdot\frac{x-2}{5(x+2)}=\frac{1}{x-1}\cdot\frac{x-2}{5}=\frac{x-2}{5x-5}\\\\D:x eq1\ \wedge\ x eq2\ \wedge\ x eq-2\)