Ta có : 

1/6 < 1/5 , 1/7 < 1/5 , ... 1/19 < 1/5

=> 1/6 + 1/7 + ...+ 1/19 < 1/5 + 1/5 + ...+ 1/5

=> 1/6 + 1/7 + ...+ 1/19  < 1/5 . 14 

=> 1/6 + 1/7 + ...+ 1/19 < 14/5 = 2 , 8 

Đặt A = \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+....+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}\) 

\(A=\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}\right)+\left(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{14}\right)+\left(\frac{1}{15}+\frac{1}{16}+...+\frac{1}{19}\right)\) 

\(\Rightarrow A< \left(\frac{1}{5}+...+\frac{1}{5}\right)+\left(\frac{1}{10}+...+\frac{1}{10}\right)+\left(\frac{1}{15}+...+\frac{1}{15}\right)\)

\(\Rightarrow A< \frac{1}{5}\cdot5+\frac{1}{10}\cdot5+\frac{1}{15}\cdot5\)

\(\Rightarrow A< 1+\frac{1}{2}+\frac{1}{3}\)

\(\Rightarrow A< \frac{11}{6}< 2\) 

\(\Rightarrow A< 2\left(đpcm\right)\)

\(A=\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+...+\frac{1}{20}\)

\(=\left(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}\right)+\frac{1}{12}+\left(\frac{1}{13}+...+\frac{1}{16}\right)+\left(\frac{1}{17}+...+\frac{1}{20}\right)\)

\(>\left(\frac{1}{9}+\frac{1}{9}+\frac{1}{9}\right)+\left(\frac{1}{12}+\frac{1}{12}+\frac{1}{12}\right)+\frac{1}{12}+\left(\frac{1}{16}+...+\frac{1}{16}\right)+\left(\frac{1}{24}+...+\frac{1}{24}\right)\)

\(=\frac{1}{3}+\frac{1}{4}+\frac{1}{12}+\frac{1}{4}+\frac{1}{6}=1+\frac{1}{12}\)

\(B=\frac{1}{5}+\frac{1}{6}+...+\frac{1}{18}+\frac{1}{19}\)

\(=\left(\frac{1}{5}+...+\frac{1}{9}\right)+\left(\frac{1}{10}+...+\frac{1}{14}\right)+\left(\frac{1}{15}+...+\frac{1}{19}\right)\)

\(< \left(\frac{1}{5}+...+\frac{1}{5}\right)+\left(\frac{1}{10}+...+\frac{1}{10}\right)+\left(\frac{1}{15}+...+\frac{1}{15}\right)\)

\(=\frac{5}{5}+\frac{5}{10}+\frac{5}{15}=1+\frac{5}{6}\)

Ta có :

\(S=\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+.......+\frac{1}{62}+\frac{1}{63}+\frac{1}{64}\)

\(\Rightarrow S< \frac{1}{17}+\frac{1}{17}+......+\frac{1}{17}+\frac{1}{17}+\frac{1}{17}\)

\(\Rightarrow S< \frac{1}{17}.48\)

\(\Rightarrow S< \frac{48}{17}\)

\(\Rightarrow S< 2\)( 1 ) 

Lại có :

\(S>\frac{1}{64}+\frac{1}{64}+.........+\frac{1}{64}+\frac{1}{64}+\frac{1}{64}\)

\(\Rightarrow S>\frac{1}{64}.48\)

\(\Rightarrow S>\frac{3}{4}\)( 2 ) 

Từ ( 1 ) và ( 2 ) suy ra : \(\frac{3}{4}< S< 2\)

Vậy \(1< S< 2\left(ĐPCM\right)\)

-5 phan14 và 30 phân -84 có bằng nhau không tại sao

chào anh coolkid em xin phép giải

ta nhân 17 với vế trái ta có:

\(17VT=17\cdot\frac{\left(17^{18}+1\right)}{17^{19}+1}\)

\(17VT=\frac{17^{19}+17}{17^{19}+1}\)

\(17VT=\frac{\left(17^{19}+1\right)+16}{17^{19}+1}\)

=> \(17VT=1+\frac{16}{17^{19}+1}\)

ta rút gọn biểu thức ở vế phải

\(\Rightarrow VP=\frac{\left(17^{18}+17\right)}{17^{19}+17}=\frac{17\cdot\left(17^{17}+1\right)}{17\cdot\left(17^{18}+1\right)}=\frac{17^{17}+1}{17^{18}+1}\)

\(\Rightarrow17VP=17\cdot\frac{17^{17}+1}{17^{18}+1}\)

\(17VP=\frac{\left(17^{18}+17\right)}{17^{18}+1}\)

\(17VP=1+\frac{16}{17^{18}+1}\)

mà ta có \(17^{19}+1>17^{18}+1\)

=> \(\frac{16}{17^{19}+1}<\frac{16}{17^{18}+1}\)

=> \(17VT<17VP\)

=> \(VT<VP\) (đpcm)

bạn ơi đề sai ở chỗ dấu "  ,  "  phải không?? bạn hãy sửa đề đi 

Bạn Nguyễn Thị Bích Phương ơi, mình sửa lại đề rồi đó. Bạn giải giúp mình với.