\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}{\dfrac{19}{1}+\dfrac{18}{2}+\dfrac{17}{3}+....+\dfrac{1}{19}}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}{1+\left(\dfrac{18}{2}+1\right)+\left(\dfrac{17}{3}+1\right)+\left(\dfrac{1}{19}+1\right)}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}{1+\dfrac{20}{2}+\dfrac{20}{3}+...+\dfrac{20}{19}}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}{20.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}\right)}\)

\(=\dfrac{1}{20}\)

\(\frac{1}{19}+\frac{2}{18}+\frac{3}{17}+...+\frac{18}{2}+\frac{19}{1}\) = \(\left(\frac{1}{19}+1\right)+\left(\frac{2}{18}+1\right)+...+\left(\frac{18}{2}+1\right)+1\)

= \(\frac{20}{19}+\frac{20}{18}+...+\frac{20}{2}+\frac{20}{20}\)

=\(20.\left(\frac{1}{19}+\frac{1}{18}+...+\frac{1}{2}+\frac{1}{20}\right)\)

=\(20.\left(\frac{1}{20}+\frac{1}{19}+\frac{1}{18}+...+\frac{1}{2}\right)\)  

Vì tử số gấp 20 lần mẫu số nên phân số này bằng 20

\(A=\dfrac{1}{20}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{19\cdot20}\right)\)

\(=\dfrac{1}{20}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\)

=1/20-19/20=-18/20=-9/10

bạn viết vậy khó hiểu quá bạn viết bằng kí tự phân số ik ạ

\(A=\dfrac{19+\dfrac{18}{2}+\dfrac{17}{3}+\dfrac{16}{4}+...+\dfrac{1}{19}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}\)

Biến đổi tử số 

\(19+\dfrac{18}{2}+\dfrac{17}{3}+\dfrac{16}{4}+...+\dfrac{1}{19}\)

= 1 + \(\left(1+\dfrac{18}{2}\right)+\left(1+\dfrac{17}{3}\right)+\left(1+\dfrac{16}{4}\right)+...+\left(1+\dfrac{1}{19}\right)\)

= \(\dfrac{20}{20}+\dfrac{20}{2}+\dfrac{20}{3}+...+\dfrac{1}{19}\)

= 20 x \(\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}\right)\)

Vậy \(A=\dfrac{19+\dfrac{18}{2}+\dfrac{17}{3}+\dfrac{16}{4}+...+\dfrac{1}{19}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}\)

= \(\dfrac{20\times\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}=20\)

Vậy A = 20

c.ơn nhìu a

\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{20}\)

\(\Rightarrow\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+..+\frac{1}{20}\left(19SH\right)\)

\(\Rightarrow\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+..+\frac{1}{20}>\frac{19}{20}\)

Vậy ................

Đặt \(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{20}\) ta có : 

\(A>\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\)

Do có \(20-2+1=19\) phân số \(\frac{1}{20}\) nên : 

\(A>19.\frac{1}{20}=\frac{19}{20}\)

Vậy \(A>\frac{19}{20}\)

Chúc bạn học tốt ~ 

* Cách làm : Tử giữ nguyên,còn mẫu ta biến đổi như sau:
Mẫu : ( \(\frac{19}{1}\)+ 1 ) + ( \(\frac{18}{2}\)+ 1 ) + ( \(\frac{17}{3}\)+ 1 ) +...+ ( \(\frac{3}{17}\)+ 1 ) + ( \(\frac{2}{18}\)+ 1 ) + ( \(\frac{1}{19}\)+ 1 ) - 19  ( vì ta cộng với 19 số 1 nên phải trừ 19 )
= \(\frac{20}{1}\)+  \(\frac{20}{2}\)+  \(\frac{20}{3}\)+...+  \(\frac{20}{17}\)+  \(\frac{20}{18}\)+  \(\frac{20}{19}\)- 19
=  \(\frac{20}{2}\)+  \(\frac{20}{3}\)+...+  \(\frac{20}{17}\)+   \(\frac{20}{18}\)+  \(\frac{20}{19}\)+ ( \(\frac{20}{1}\)- 19)
=  \(\frac{20}{2}\)+  \(\frac{20}{3}\)+ ...+   \(\frac{20}{17}\)+  \(\frac{20}{18}\)+  \(\frac{20}{19}\)+  \(\frac{20}{20}\)
= 20.( \(\frac{1}{2}\)+  \(\frac{1}{3}\)+...+  \(\frac{1}{17}\)+  \(\frac{1}{18}\)+  \(\frac{1}{19}\)+  \(\frac{1}{20}\))
=> \(\frac{Tử}{Mâu}\)=  \(\frac{1}{20}\)

Phùng Quang Thịnh biến đổi sai 1 chỗ kìa 

-19 = \(\frac{20}{20}-20\)chứ mà bạn

thank Lê Tài Bảo Châu nhá