Gọi các phân số cần tìm là: \(\dfrac{a}{b}\) theo bài ra ta có:

                     \(\dfrac{a}{b}\) =  \(\dfrac{a+2}{b\times2}\) 

            a.(b x 2) = (a + 2) x b

              ab x 2 = ab + 2b

                   ab = 2b

                   a = 2

                 Ta có: \(\dfrac{2}{b}\) > \(\dfrac{1}{5}\) = \(\dfrac{2}{10}\)

             ⇒ b < 10 ⇒ b = 1; 2; 3; 4; 5; 6; 7; 8; 9

Vì \(\dfrac{2}{b}\) không phải là số tự nhiên nên b \(\in\) {3; 4; 5; 6; 7; 8; 9}

Bài 16:

\(\dfrac{1}{6}\) < \(\dfrac{1}{5^2}\) + \(\dfrac{1}{6^2}\) + \(\dfrac{1}{7^2}\) +...+ \(\dfrac{1}{100^2}\) < \(\dfrac{1}{4}\)

\(\dfrac{1}{5^2}\) < \(\dfrac{1}{4.5}\) = \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)

\(\dfrac{1}{6^2}\) < \(\dfrac{1}{5.6}\) = \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\)

............................

\(\dfrac{1}{100^2}\) < \(\dfrac{1}{99.100}\) = \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\)

Cộng vế với vế ta có: 

\(\dfrac{1}{5^2}\) + \(\dfrac{1}{6^2}\)+...+ \(\dfrac{1}{100^2}\) < \(\dfrac{1}{4}\) - \(\dfrac{1}{100}\) < \(\dfrac{1}{4}\) (1)

\(\dfrac{1}{5^2}\) > \(\dfrac{1}{5.6}\) = \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\)

\(\dfrac{1}{6^2}\) > \(\dfrac{1}{6.7}\) = \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\)

...............................

\(\dfrac{1}{100^2}\) > \(\dfrac{1}{100.101}\) = \(\dfrac{1}{100}\) - \(\dfrac{1}{101}\)

Cộng vế với vế ta có:

\(\dfrac{1}{5^2}\) + \(\dfrac{1}{6^2}\) + ... + \(\dfrac{1}{100^2}\) > \(\dfrac{1}{5}\) - \(\dfrac{1}{101}\)= \(\dfrac{96}{505}\) > \(\dfrac{96}{576}\) = \(\dfrac{1}{6}\) (2)

Kết hợp (1) và (2) ta có: 

\(\dfrac{1}{6}\) < \(\dfrac{1}{5^2}\) + \(\dfrac{1}{6^2}\) +...+ \(\dfrac{1}{100^2}\) < \(\dfrac{1}{4}\) (đpcm)

\(A=\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^6}+\dots+\dfrac{1}{2^{100}}\)

\(2A=\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dots+\dfrac{1}{2^{99}}\)

\(2A-A=\left(\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dots+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^6}+\dots+\dfrac{1}{2^{100}}\right)\)

\(A=\dfrac{1}{2^3}-\dfrac{1}{2^{100}}\)

\(A=\dfrac{2^{97}}{2^{100}}-\dfrac{1}{2^{100}}\)

\(A=\dfrac{2^{97}-1}{2^{100}}\)