(1-$\frac{1}{2^{2}}$).(1-$\frac{1}{3^{2}}$).(1-$\frac{1}{4^{2}}$)…(1-$\frac{1}{10^{2}}$)

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1. Đáp án:

    $\frac{11}{20}$

   Giải thích các bước giải:

   $(1-\frac{1}{2^2}).(1-\frac{1}{3^2}).(1-\frac{1}{4^2})…(1-\frac{1}{10^2})\\
   =\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}…\frac{10^2-1}{10^2}\\
   =\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}….\frac{99}{10^2}\\
   =\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}….\frac{11.9}{10^2}\\
   =\frac{1.2.3….11}{2.3.4.5…10}.\frac{3.4.5…9}{2.3.4.5…10}\\
   =\frac{11}{1}.\frac{1}{2.10}\\
   =\frac{11}{20}$

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