Question

45.(7373/3030+7373/4242+7373/5656+7373/7272)

Answer 1

\(45\left(\frac{7373}{3030}+\frac{7373}{4242}+\frac{7373}{5656}+\frac{7373}{7272}\right)\)

\(=45\left(\frac{73.101}{30.101}+\frac{73.101}{42.101}+\frac{73.101}{56.101}+\frac{73.101}{72.101}\right)\)

\(=45\left(\frac{73}{30}+\frac{73}{42}+\frac{73}{56}+\frac{73}{42}\right)\)

\(=45.73\left(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)

\(=45.73\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)

\(=45.73\left(\frac{1}{5}-\frac{1}{9}\right)\)

\(=45.73\left(\frac{9}{45}-\frac{5}{45}\right)\)

\(=45.73.\frac{4}{45}\)

\(=292\)