Đáp án:
$\frac{5}{9}$
Giải thích các bước giải:
$\left ( 1-\frac{1}{2}^2 \right ).\left ( 1-\frac{1}{3}^2 \right ).\left ( 1-\frac{1}{4}^2 \right )...\left ( 1-\frac{1}{9}^2 \right )\\
=\left ( 1-\frac{1}{4} \right ).\left ( 1-\frac{1}{9} \right ).\left ( 1-\frac{1}{16} \right )...\left ( 1-\frac{1}{81} \right )\\
=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{80}{81}\\
=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}....\frac{8.10}{9^2}\\
=\frac{1.2.3...8}{2.3.4.5...9}.\frac{3.4.5...10}{2.3.4.5....9}\\
=\frac{1}{9}.\frac{10}{2}\\
=\frac{10}{18}=\frac{5}{9}$