Lời giải:
\(S=\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{9^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8..9}\)
hay \(S< \frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{9-8}{8.9}\)
\(S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
\(S< 1-\frac{1}{9}=\frac{8}{9}\) (1)
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Mặt khác:
\(S> \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\\ S> \frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{10-9}{9.10}\\ S> \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\\ S> \frac{1}{2}-\frac{1}{10}=\frac{2}{5}(2)\)
Từ $(1); (2)$ ta có đpcm
2*2 là 2 x 2 hay 22 thế em?