\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)

\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+...+\dfrac{19}{81.100}\)\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{81}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}< 1\)

\(A=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)

\(A=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}\)

\(A=\dfrac{2^2}{1^2.2^2}-\dfrac{1^2}{1^2.2^2}+\dfrac{3^2}{2^2.3^2}-\dfrac{2^2}{2^2.3^2}+...+\dfrac{10^2}{9^2.10^2}-\dfrac{9^2}{9^2.10^2}\)\(A=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)

\(A=1-\dfrac{1}{10^2}< 1\left(đpcm\right)\)

$A=\frac{3}{1^2{.2}^2}+\frac{5}{2^2{.3}^2}+\frac{7}{3^2{.4}^2}+...+\frac{19}{9^2{.10}^2}$

A=\(\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+....+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)

A = \(\dfrac{1}{1^2}-\dfrac{1}{10^2}\)

A = \(1-\dfrac{1}{10^2}\) < 1

Vậy A < 1

\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2+10^2}\)

\(=\left(\dfrac{1}{1^2}-\dfrac{1}{2^2}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{3^2}\right)+...+\left(\dfrac{1}{9^2}-\dfrac{1}{10^2}\right)\)

=\(\dfrac{1}{1}-\dfrac{1}{10^2}\)

\(=1-\dfrac{1}{100}\)

Mà \(1-\dfrac{1}{100}< 1\)

\(\Rightarrow\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+...+\dfrac{19}{9^2.10^2}< 1\) (đpcm)

\(D=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)

\(D=\left(\dfrac{1}{1^2}-\dfrac{1}{2^2}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{4^2}\right)+...+\left(\dfrac{1}{9^2}-\dfrac{1}{10^2}\right)\)

\(D=\dfrac{1}{1}-\dfrac{1}{10^2}\)

\(D=1-\dfrac{1}{100}< 1\)

Vậy \(D< 1\left(đpcm\right)\)

Bạn tham khảo lời giải tại đây:

https://olm.vn/hoi-dap/detail/81621153379.html

\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)

\(=\left(\dfrac{1}{1^2}-\dfrac{1}{2^2}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^2}-\dfrac{1}{4^2}\right)+...+\left(\dfrac{1}{9^2}-\dfrac{1}{100^2}\right)\)

\(=\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)

\(=\dfrac{1}{1}-\dfrac{1}{100}\)

\(=\dfrac{99}{100}\)

\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+...+\dfrac{19}{9^2.10^2}\)

\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+...+\dfrac{19}{81.100}\)

\(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+...+\dfrac{1}{81}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)

Ta có:

\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)

= \(\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}\)

= \(\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)

= \(1-\dfrac{1}{10^2}\)

Mà \(1-\dfrac{1}{10^2}< 1\) nên:

\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\) < 1 (đpcm).