có nhầm đề không vậy phải là 2010-

minh lam duoc roi . cach viet phan so ban bam vao o mau vang o cuoi trang .cu di con chuot xuong cuoi trang thi thay 1 o vang , vao xem huong dan la biet ngay ma.

Gọi a là tử số, b là mẫu số của phân số A

a = \(\frac{2008}{1}\)+ \(\frac{2007}{2}\)+ \(\frac{2006}{3}\)+ ... + \(\frac{1}{2008}\)

Dãy số a có (2008 - 1)  : 1 + 1 = 2008 số. Và a = ( \(\frac{2008}{1}\)+ \(\frac{1}{2008}\)) x (2008 : 2) 

b = \(\frac{1}{2}\)+ \(\frac{1}{3}\)+ \(\frac{1}{4}\)+ ... + \(\frac{1}{2009}\)

Dãy số b có (2009 - 2) : 1 + 1 = 2008 số. Và b = (\(\frac{1}{2}\)+ \(\frac{1}{2009}\)) x (2008 : 2)

A = [ ( \(\frac{2008}{1}\)+ \(\frac{1}{2008}\)) x (2008 : 2)] : [ (\(\frac{1}{2}\)+ \(\frac{1}{2009}\)) x (2008 : 2)] = ( \(\frac{2008}{1}\)+ \(\frac{1}{2008}\)) :  (\(\frac{1}{2}\)+ \(\frac{1}{2009}\)) 

A = \(\frac{\text{2008 x2008 + 1}}{2008}\)x \(\frac{2x2009+2}{2x2009}\)

A = 2008

Xét tử ta có:

\(2008+\frac{2007}{2}+\frac{2006}{3}+....+\frac{1}{2008}\)

= \(1+\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+...+\left(1+\frac{1}{2008}\right)\)

= \(\frac{2009}{2009}+\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2008}\)

= \(2009.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}\right)\)

=> A = \(\frac{2009.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}}\)

=> A = 2009

A=\(\frac{\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+\left(1+\frac{2005}{4}\right)+...........+\left(1+\frac{2}{2008}\right)+\left(1+\frac{1}{2009}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{2008}+\frac{1}{2009}}\)=\(\frac{\frac{2009}{2}+\frac{2009}{3}+\frac{2009}{4}+....+\frac{2009}{2008}+\frac{2009}{2009}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2008}+\frac{1}{2009}}\frac{ }{ }\)  

                                                                                                               =\(\frac{2009\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2008}+\frac{1}{2009}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2008}+\frac{1}{2009}}\frac{ }{ }\) 

                                                                                                                =2009 

Vay A=2009

2008 + 2007/2 + 2006/3 + 2005/4 + ... + 2/2007 + 1/2008

2009-1/1 + 2009-2/2 + 2009-3/3 + 2009-4/4 + ... + 2009-2007/2007 + 2009-2008/2008

2009 - 1 + 2009/2 - 1 + 2009/3 - 1 + 2009/4 - 1 + ... + 2009/2007 - 1 + 2009/2008 - 1

2009 + 2009.(1/2 + 1/3 + 1/4 + ... + 1/2007 + 1/2008 ) - ( 1 + 1 + 1 + 1 + ... + 1 + 1 )

2009 + 2009.( 1/2 + 1/3 + 1/4 + ... + 1/2007 + 1/2008 ) - 2008

1 + 2009.( 1/2 + 1/3 + 1/4 + ... + 1/2007 + 1/2008 )

2009.( 1/2 + 1/3 + 1/4 + ... + 1/2007 + 1/2008 + 1/2009 )

=> giá trị của biểu thức trên là 2009

2009 chắc chắn tính rồi

\(A=\frac{2008+\frac{2007}{2}+\frac{2006}{3}+\frac{2005}{4}+...+\frac{2}{2007}+\frac{1}{2008}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}\)

\(=\frac{\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+\left(1+\frac{2005}{4}\right)+...+\left(1+\frac{1}{2007}\right)+\left(1+\frac{1}{2008}\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}\)

\(=\frac{\frac{2009}{2}+\frac{2009}{3}+\frac{2009}{4}+...+\frac{2009}{2007}+\frac{2009}{2008}+\frac{2009}{2009}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}\)

\(=\frac{2009\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}=2009\)

$=\frac{2008+\frac{2007}{2}+\frac{2006}{3}+\frac{2005}{4}+...+\frac{2}{2007}+\frac{1}{2008}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}$

$1+\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+...+\left(1+\frac{1}{2008}\right)$

$\frac{2009}{2009}+\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2008}$

$2009.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}\right)$

A=$\frac{2009.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}}$

A=2009

bằng 2009 nha