tính rút gọn

49) 
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\left( 1-\frac{1}{3^{2}}\right)\left( 1-\frac{1}{4^{2}}\right)\left( 1-\frac{1}{5^{2}}\right) .......\left( 1-\frac{1}{999^{2}}\right)\\
=\left( 1-\frac{1}{3}\right)\left( 1+\frac{1}{3}\right)\left( 1-\frac{1}{4}\right)\left( 1+\frac{1}{4}\right)\left( 1-\frac{1}{5}\right)\left( 1+\frac{1}{5}\right) ........\left( 1-\frac{1}{999}\right)\left( 1+\frac{1}{999}\right)\\
=\left[\left( 1-\frac{1}{3}\right)\left( 1-\frac{1}{4}\right)\left( 1-\frac{1}{5}\right) ....\left( 1-\frac{1}{999}\right)\right]\left[\left( 1+\frac{1}{3}\right)\left( 1+\frac{1}{4}\right)\left( 1+\frac{1}{5}\right) ....\left( 1+\frac{1}{999}\right)\right]\\
=\left(\frac{2}{3} .\frac{3}{4} .\frac{4}{5} .....\frac{998}{999}\right) .\left(\frac{4}{3} .\frac{5}{4} .\frac{6}{5} .....\frac{1000}{999}\right)\\
=\frac{2}{999} .\frac{1000}{3} =\frac{2000}{2997}
\end{array}$

50) 
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\left( 1-\frac{1}{4^{2}}\right)\left( 1-\frac{1}{5^{2}}\right)\left( 1-\frac{1}{6^{2}}\right) .......\left( 1-\frac{1}{2999^{2}}\right)\\
=\left( 1-\frac{1}{4}\right)\left( 1+\frac{1}{4}\right)\left( 1-\frac{1}{5}\right)\left( 1+\frac{1}{5}\right)\left( 1-\frac{1}{6}\right)\left( 1+\frac{1}{6}\right) ........\left( 1-\frac{1}{2999}\right)\left( 1+\frac{1}{2999}\right)\\
=\left[\left( 1-\frac{1}{4}\right)\left( 1-\frac{1}{5}\right)\left( 1-\frac{1}{6}\right) ....\left( 1-\frac{1}{2999}\right)\right]\left[\left( 1+\frac{1}{4}\right)\left( 1+\frac{1}{5}\right)\left( 1+\frac{1}{6}\right) ....\left( 1+\frac{1}{2999}\right)\right]\\
=\left(\frac{3}{4} .\frac{4}{5} .\frac{5}{6} .....\frac{2998}{2999}\right) .\left(\frac{5}{4} .\frac{6}{5} .\frac{7}{6} .....\frac{3000}{2999}\right)\\
=\frac{3}{2999} .\frac{3000}{4} =\frac{2250}{2999}
\end{array}$