Cho (A = left( {frac{1}{{{2^2}}} - 1} ight).left( {frac{1}{{{3^2}}} - 1} ight).left( {frac{1}{{{4^2}}} - 1} ight)...

* Hướng dẫn giải

Ta có A là tích của 99 số âm nên A < 0. Do đó:

\(\begin{array}{l}
A =  - \left[ {\left( {1 - \frac{1}{{{2^2}}}} \right).\left( {1 - \frac{1}{{{3^2}}}} \right).\left( {1 - \frac{1}{{{4^2}}}} \right)...\left( {1 - \frac{1}{{{{100}^2}}}} \right)} \right]\\
A =  - \left[ {\left( {1 - \frac{1}{4}} \right).\left( {1 - \frac{1}{9}} \right).\left( {1 - \frac{1}{{16}}} \right)...\left( {1 - \frac{1}{{10000}}} \right)} \right]\\
A =  - \left( {\frac{3}{{{2^2}}}} \right..\frac{8}{{{3^2}}}.\frac{{15}}{{{4^2}}}...\left. {\frac{{9999}}{{{{100}^2}}}} \right)\\
A =  - \left( {\frac{{1.3}}{{{2^2}}}} \right..\frac{{2.4}}{{{3^2}}}.\frac{{3.5}}{{{4^2}}}...\left. {\frac{{99.101}}{{{{100}^2}}}} \right)\\
A =  - \left( {\frac{{1.2.3...98.99}}{{2.3.4...99.100}}} \right..\left. {\frac{{3.4.5...100.101}}{{2.3.4...99.100}}} \right)\\
A =  - \left( {\frac{1}{{100}}} \right..\left. {\frac{{101}}{2}} \right) =  - \frac{{101}}{{200}} <  - \frac{1}{2}
\end{array}\)

Vậy \(A <  - \frac{1}{2}\)