Đáp án:
$\frac{11}{20}$
Giải thích các bước giải:
$(1-\frac{1}{2^2}).(1-\frac{1}{3^2}).(1-\frac{1}{4^2})...(1-\frac{1}{10^2})\\
=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}...\frac{10^2-1}{10^2}\\
=\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}....\frac{99}{10^2}\\
=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}....\frac{11.9}{10^2}\\
=\frac{1.2.3....11}{2.3.4.5...10}.\frac{3.4.5...9}{2.3.4.5...10}\\
=\frac{11}{1}.\frac{1}{2.10}\\
=\frac{11}{20}$
