\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\)

\(B=\frac{2}{1004.2006}+\frac{2}{1005.2005}+...+\frac{1}{1505.1505}\)

tới đây tự làm

http://pitago.vn/question/cho-a-1-1-2-1-3-4-1-2003-2004-1-2005-2006-va-b-46191.html tham khảo đi

A=\(\frac{1}{1.2}+....+\frac{1}{2013.2014}+\frac{1}{2015.2016}\)=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2015}-\frac{1}{2016}\)

=\(\left(1+\frac{1}{3}+.....+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2016}\right)\)

=\(\left(1+\frac{1}{2}+.....+\frac{1}{2016}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2016}\right)\)

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{2005.2006}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2005}-\frac{1}{2006}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{2005}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)

\(=\left(1+\frac{1}{2}+...+\frac{1}{2006}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)

\(=\left(1+\frac{1}{2}+...+\frac{1}{2006}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1003}\right)\)

\(=\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}\)(1)

\(B=\frac{1}{1004.2006}+\frac{1}{1005.2005}+....+\frac{1}{2006.1004}\)

\(\Rightarrow\frac{1}{1004}+\frac{1}{2006}+\frac{1}{1005}+\frac{1}{2005}+...+\frac{1}{2006}+\frac{1}{1004}=2\left(\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}\right)\)

\(=\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{1505}\)(2)

Thế (1) và (2) vào ta có:

\(\frac{A}{B}=\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{1505}}\)

A=2005/2006

A=1−12+13−14+...+12005−12006=(1+12+...+12006)−(1+12+..+11003)=11004+11005+...+12006" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:41.489em; padding:1px 0px; position:relative; white-space:nowrap; width:749.281px; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">A=1−12+13−14+...+12005−12006=(1+12+...+12006)−(1+12+..+11003)=11004+11005+...+12006

13010.B=11004+12006+11005+12005+...+11004=11505(11004+11005+...+12006)" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">13010.B=11004+12006+11005+12005+...+11004=11505(11004+11005+...+12006)

Suy ra A/B = 1505

Tham khảo nha 

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2005.2006}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2005}-\frac{1}{2006}\)

\(=1-\frac{1}{2006}\)

\(=\frac{2005}{2006}\)

\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2003.2004}+\frac{1}{2005.2006}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2005}-\frac{1}{2006}\)\(=\left(1+\frac{1}{3}+...+\frac{1}{2005}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)

\(=\left(1+\frac{1}{2}+...+\frac{1}{2006}\right)-2\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)\(=\left(1+\frac{1}{2}+...+\frac{1}{2006}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1003}\right)=\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}\)

\(B=\frac{1}{1004.2006}+\frac{1}{1005.2005}+\frac{1}{1006.2004}+...+\frac{1}{2006.1004}\)

=>3010B=\(\frac{1}{1004}+\frac{1}{2006}+\frac{1}{1005}+\frac{1}{2005}+...+\frac{1}{2006}+\frac{1}{1004}=2\cdot\left(\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}\right)\)

=>B=\(\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{1505}\)

=>\(\frac{A}{B}=\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{1505}}=1505\)

Kết quả là \(1505\)

K nha!

là 1505

\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{99.100}\)

\(B=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)

\(B=\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{100}\right)\)

\(B=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(B=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{25}\right)\)

\(B=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+....+\frac{1}{100}=A\)

=>A/B=1

B=\(\frac{1}{1004.2006}+\frac{1}{1005.2005}+...+\frac{1}{2006.1004}\)

                                                                     BÀI GIẢI

A=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2005}-\frac{1}{2006}=\left(1+\frac{1}{2}+...+\frac{1}{2006}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1003}\right)\)

  =\(\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}\)

Lại có \(\frac{1}{3010}.B=\frac{1}{1004}+\frac{1}{2006}+\frac{1}{1005}+\frac{1}{2005}+...+\frac{1}{1004}=1505.\left(\frac{1}{1004}+...+\frac{1}{2006}\right)\)

Vậy A/B=1505. Từ bài toán này, chắc cx nghĩ ra cách làm rồi nhỉ

BẤM ĐÚNG CHO TUI

Mình mở rộng bài toán nhé, xong tự nghĩ cách giải . Đề mở rộng là:

Tính A/B biết \(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2005.2006}\)