\(\dfrac{1212}{1515}+\dfrac{1212}{3535}+\dfrac{1212}{6363}+\dfrac{1212}{9999}\)
=\(\dfrac{12}{15}+\dfrac{12}{35}+\dfrac{12}{63}+\dfrac{12}{99}\)
=\(\dfrac{16}{11}\)
Giải:
\(\dfrac{1212}{1515}+\dfrac{1212}{3535}+\dfrac{1212}{6363}+\dfrac{1212}{9999}\)
\(=\dfrac{12}{15}+\dfrac{12}{35}+\dfrac{12}{63}+\dfrac{12}{99}\)
\(=12.\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}\right)\)
\(=12.\dfrac{1}{2}.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\right)\)
\(=6.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)\)
\(=6.\left(\dfrac{1}{3}-\dfrac{1}{11}\right)\)
\(=6.\dfrac{8}{33}\)
\(=\dfrac{16}{11}\)
\(=\dfrac{12}{15}+\dfrac{12}{35}+\dfrac{12}{63}+\dfrac{12}{99}=12\cdot\left[\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}\right]=12\cdot\left[\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}+\dfrac{1}{9\cdot11}\right]=12\cdot\dfrac{1}{2}\cdot\left[\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right]=6\cdot\left[\dfrac{1}{3}-\dfrac{1}{11}\right]=6\cdot\dfrac{8}{33}=\dfrac{48}{33}=\dfrac{16}{11}\)
