Ta có: \(A=\frac{\left(1+\frac{2017}{1}\right)\left(1+\frac{2017}{2}\right)...\left(1+\frac{2017}{1009}\right)}{\left(1+\frac{1009}{1}\right)\left(1+\frac{1009}{2}\right)...\left(1+\frac{1009}{2017}\right)}=\frac{\frac{2017+1}{1}\frac{2017+2}{2}...\frac{2017+1009}{1009}}{\frac{1009+1}{1}\frac{1009+2}{2}...\frac{1009+2017}{2017}}\)

\(\Leftrightarrow A=\frac{\frac{2018.2019...3026}{1.2...1009}}{\frac{1010.1011...3026}{1.2...2017}}=\frac{2018.2019...3026}{1.2...1009}.\frac{1.2...2017}{1010.1011...3026}\)

\(\Leftrightarrow A=\frac{1.2...2017.2018.2019...3026}{1.2...1009.1010.1011...3026}=\frac{1.2.3...3026}{1.2.3...3026}=1.\)

\(A=1\)

Kết quả =1

Kết quả =1

Kết quả =1

A=

).23.7.1009
1009
1
.
7
1
.
23
1
1009
1
7
1
23
1
(

).23.7.1009
1009
1
7
1
23
1
(
  
 

+
(23 7).1009 161 1

1
  

=
7.1009 23.1009 23.7 1
7.1009 23.1009 23.7
  
 
+
23.1009 7.1009 23.7 1

1
  
= 1

Ta có : \(2018.\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019.\left(\frac{1}{2017}-2\right)=\frac{2018}{2017}-2019.2-\frac{2019}{2017}+2019.2\)

\(=\frac{2018}{2017}-\frac{2019}{2017}=-\frac{1}{2017}\)

\(2018.\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019.\left(\frac{1}{2017}-2\right)\)

\(=\frac{2018}{2017}-2018.\frac{2019}{1009}-\frac{2019}{2017}+2019.2\)

\(=\frac{2018}{2017}-2.2019-\frac{2019}{2017}+2.2019\)

\(=\frac{2018}{2017}-\frac{2019}{2017}=-\frac{1}{2017}\)

\(2018\cdot\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019\cdot\left(\frac{1}{2017}-2\right)=\frac{2018}{2017}-4038-\frac{2019}{2017}+4038\)

\(=\frac{2018}{2017}-\frac{2019}{2017}=-\frac{1}{2017}\)

=> A= \(\frac{\left(\frac{1}{23}+\frac{1}{7}-\frac{1}{1009}\right).23.7.1009}{\left(\frac{1}{23}+\frac{1}{7}-\frac{1}{1009}+\frac{1}{7}.\frac{1}{23}.\frac{1}{1009}\right).23.7.1009}\) + \(\frac{1}{30.1009-160}\)

=> A= \(\frac{7.1009+23.1009-23.7}{7.1009+23.1009-23.7+1}\) + \(\frac{1}{7.1009+23.1009-23.7+1}\) = \(\frac{7.1009+23.1009-23.7+1}{7.1009+23.1009-23.7+1}\) = 1.

Hưng ơi có hiểu bài này ko bảo t với!!!!!!!!