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)

c

C. Nhường cơm sẻ áo

`#3107.101107`

`(3x - 4)^3 = 5^2 + 4*5^2`

`\Rightarrow (3x - 4)^3 = 5^2 * (1 + 4)`

`\Rightarrow (3x - 4)^3 = 5^2 * 5`

`\Rightarrow (3x - 4)^3 = 5^3`

`\Rightarrow 3x - 4 = 5`

`\Rightarrow 3x = 5 + 4`

`\Rightarrow 3x = 9`

`\Rightarrow x = 9 \div 3`

`\Rightarrow x = 3`

Vậy, `x = 3.`

\((3x-4)^3=5^2+4\cdot5^2\\\Rightarrow(3x-4)^3=5^2\cdot(1+4)\\\Rightarrow(3x-4)^3=5^2\cdot5\\\Rightarrow(3x-4)^3=5^3\\\Rightarrow3x-4=5\\\Rightarrow3x=5+4\\\Rightarrow3x=9\\\Rightarrow x=9:3\\\Rightarrow x=3\)

`#3107.101107`

\(\dfrac{4}{-5}\cdot\dfrac{8}{17}-\dfrac{4}{5}\div\dfrac{17}{9}+1\dfrac{4}{5}\)

\(=-\dfrac{4}{5}\cdot\dfrac{8}{17}-\dfrac{4}{5}\cdot\dfrac{9}{17}+1+\dfrac{4}{5}\)

\(=\dfrac{4}{5}\cdot\left(-\dfrac{8}{17}-\dfrac{9}{17}+1\right)+1\)

\(=\dfrac{4}{5}\cdot\left(-1+1\right)+1\)

\(=\dfrac{4}{5}\cdot0+1\)

\(=0+1\)

\(=1\)

Tuổi của ông hiện nay:

(84 + 62) : 2 = 73 (tuổi)

Tuổi cháu là:

73 - 62 = 11 (tuổi)

dạ cảm ơn

`#3107.101107`

$20 \cdot 18 \cdot 20 \cdot 17 - 20 \cdot 17 \cdot 20 \cdot 16$

$= 20 \cdot 17 \cdot 20 \cdot (18 - 16)$

$= 20 \cdot 17 \cdot 20 \cdot 2$

$= 400 \cdot 34$

$= 13600$

\(20\times18\times20\times17-20\times17\times20\times16\)

\(=20\times20\times17\times\left(18-16\right)\)

\(=20\times20\times17\times2\)

\(=400\times34\)

\(=13600\)