$$\frac{{{{\left( {3\frac{2}{3}} \right)}^2} - {{\left( {2\frac{1}{2}} \right)}^2}}}{{{{\left( {4\frac{3}{4}} \right)}^2} - {{\left( {3\frac{1}{3}} \right)}^2}}}$$   $$ \div $$ $$\frac{{3\frac{2}{3} - 2\frac{1}{2}}}{{4\frac{3}{4} - 3\frac{1}{3}}}$$   = ?

Solution (By Examveda Team)

$$\eqalign{ & {\text{If }}a = 3\frac{2}{3}, \cr & b = 2\frac{1}{2}, \cr & c = 4\frac{3}{4}, \cr & d = 3\frac{1}{3}, \cr & {\text{then}} \cr & {\text{Given expression ,}} \cr & \,\,\,\,\frac{{\left( {{a^2} - {b^2}} \right)}}{{\left( {{c^2} - {d^2}} \right)}} \div \frac{{\left( {a - b} \right)}}{{\left( {c - d} \right)}} \cr & = \frac{{\left( {{a^2} - {b^2}} \right)}}{{\left( {{c^2} - {d^2}} \right)}} \times \frac{{\left( {c - d} \right)}}{{\left( {a - b} \right)}} \cr & = \frac{{\left( {a + b} \right)}}{{\left( {c + d} \right)}} \cr & = \frac{{3\frac{2}{3} + 2\frac{1}{2}}}{{4\frac{3}{4} + 3\frac{1}{3}}} \cr & = \frac{{\frac{{11}}{3} + \frac{5}{2}}}{{\frac{{19}}{4} + \frac{{10}}{3}}} \cr & = \frac{{37}}{6} \times \frac{{12}}{{97}} \cr & = \frac{{74}}{{97}} \cr} $$