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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
Ta có :
\(B=\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}\)
\(B=1+\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+...+\left(1+\frac{2}{2007}\right)+\left(1+\frac{1}{2008}\right)\)
\(B=\frac{2009}{2009}+\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2007}+\frac{2009}{2008}\)
\(B=2009\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}\right)\)
\(\Rightarrow\)\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}}{2009\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}\right)}=\frac{1}{2009}\)
Vậy \(\frac{A}{B}=\frac{1}{2009}\)
\(B=\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{1007}+\frac{1}{2008}\)
\(B=\frac{2008}{1}+1+\frac{2007}{2}+1+\frac{2006}{3}+1+....+\frac{2}{2007}+1+\frac{1}{2008}+1-2008\)
\(B=\frac{2009}{1}+\frac{2009}{2}+\frac{2009}{3}+.....+\frac{2009}{2007}+\frac{2009}{2008}-\frac{2009.2008}{2009}\)
\(B=2009\cdot\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2007}+\frac{1}{2008}-\frac{2008}{2009}\right)\)
Từ đó suy ra \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}}{2009\cdot\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2007}+\frac{1}{1008}+\frac{2008}{2009}\right)}\)
\(=\frac{\frac{1}{2009}}{2009\cdot\left(1+\frac{2008}{2009}\right)}\)
Bí òi
Đề của bạn sai rồi: Phải là B = \(\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}\) chứ ?!
\(A=\frac{2008+\frac{2007}{2}+\frac{2006}{3}+\frac{2005}{4}+...+\frac{2}{2007}+\frac{1}{2008}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}\)
\(=\frac{\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+\left(1+\frac{2005}{4}\right)+...+\left(1+\frac{1}{2007}\right)+\left(1+\frac{1}{2008}\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}\)
\(=\frac{\frac{2009}{2}+\frac{2009}{3}+\frac{2009}{4}+...+\frac{2009}{2007}+\frac{2009}{2008}+\frac{2009}{2009}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}\)
\(=\frac{2009\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}=2009\)
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
tách số 2008 thành 2008 số 1(=1+1+...+1),sau đó cộng vào 2007 phân số kia, mỗi phân số công thêm 1,ta dc một biểu thức tư đều lan 2009(còn thừa một số 1 các bạn hãy viết nó dưới dạng\(\frac{2009}{2009}\)lúc đó ta dc:A=\(\frac{\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2008}+\frac{2009}{2009}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}}\)
và cuối cùng ta rút gọn!có gì chưa hiu nhắn tin lại nhé!
lần sau bảo cô ra đề khó thêm:):):)
Ta co:\(B=\frac{2008}{1}+\frac{2007}{2}+...+\frac{2}{2007}+\frac{1}{2008}\)
\(B=\frac{2009-1}{1}+\frac{2009-2}{2}+...+\frac{2009-2007}{2007}+\frac{2009-2008}{2008}\)
\(B=\left(\frac{2009}{1}+\frac{2009}{2}+...+\frac{2009}{2008}\right)-\left(\frac{1}{1}+\frac{2}{2}+...+\frac{2008}{2008}\right)\)
\(B=2009+2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}\right)-2008\)
\(B=1+2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}\right)\)
\(B=2009\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2008}+\frac{1}{2009}\right)\)
Vay \(\frac{A}{B}=\frac{1}{2009}\)
mik đọc nhầm đề rồi.Kết quả là 9/187
Li-ke cho mik nhé!