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

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

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

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

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

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

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

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

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

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

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

* 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

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

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

Mẫu số=19/1+18/2+17/3+...+2/18+1/19

=(1+1+1+...+1)+(18/2+17/3+...+2/18+1/19)

(19 số 1)                      (18 phân số)

=(1+18/2)+(1+17/3)+...+(1+2/18)+(1+1/19)+1

=20/2+20/3+...+20/18+20/19+20/20

=20.(1/2+1/3+...+1/18+1/19+1/20)

Phân số trên=1/20