Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Theo bài ra ta có :
\(A=\frac{2011}{1.2}+\frac{2011}{3.4}+\frac{2011}{4.5}+...+\frac{2011}{1999.2000}\)
\(\Rightarrow\frac{A}{2011}=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{1999.2000}\)
\(\Rightarrow\frac{A}{2011}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1999}-\frac{1}{2000}\)
\(\Rightarrow\frac{A}{2011}=\left(\frac{1}{1}+\frac{1}{3}+...+\frac{1}{1999}\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{2000}\right)\)
\(\Rightarrow\frac{A}{2011}=\left(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{2000}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2000}\right)\) \(-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2000}\right)\)
\(\Rightarrow\frac{A}{2011}=\left(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{2000}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2000}\right)\)
\(\Rightarrow\frac{A}{2011}=\left(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{2000}\right)-\left(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{1000}\right)\)
\(\Rightarrow\frac{A}{2011}=\frac{1}{1001}+\frac{1}{1002}+...+\frac{1}{2000}\)
\(\Rightarrow A=2011\left(\frac{1}{1001}+\frac{1}{1002}+...+\frac{1}{2000}\right)\left(1\right)\)
Ta lại có :
\(B=\frac{2012}{1001}+\frac{2012}{1002}+...+\frac{2012}{2000}\)
\(\Rightarrow B=2012\left(\frac{1}{1001}+\frac{1}{1002}+...+\frac{1}{2000}\right)\)\(\left(2\right)\)
Từ (1) và (2) => A < B
Vậy A < B
\(\Leftrightarrow2x\left(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)
Đặt \(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100};B=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)
\(\Leftrightarrow A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ \Leftrightarrow A=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\\ \Leftrightarrow A=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\\ \Leftrightarrow A=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}-1-\dfrac{1}{2}-\dfrac{1}{3}-...-\dfrac{1}{50}\\ \Leftrightarrow A=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}=B\)
\(\Leftrightarrow2x.A=B\Leftrightarrow2x.B-B=0\\ \Leftrightarrow B\left(2x-1\right)=0\\ \Leftrightarrow2x-1=0\Leftrightarrow x=\dfrac{1}{2}\)
Ta có :
\(\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2010}+...+\frac{x-2012}{1}=2012\)
\(\Leftrightarrow\)\(\left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+\left(\frac{x-3}{2010}-1\right)+...+\left(\frac{x-2012}{1}-1\right)=2012-2012\)
\(\Leftrightarrow\)\(\frac{x-2013}{2012}+\frac{x-2013}{2011}+\frac{x-2013}{2010}+...+\frac{x-2013}{1}=0\)
\(\Leftrightarrow\)\(\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+\frac{1}{1}\right)=0\)
Vì \(\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+1\right)\ne0\)
\(\Rightarrow\)\(x-2013=0\)
\(\Rightarrow\)\(x=2013\)
Vậy \(x=2013\)
Chúc bạn học tốt
Ta có:
\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\)
=> \(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right).x=\frac{2012}{51}+\frac{2012}{52}+...+\frac{2012}{99}+\frac{2012}{100}\)
=> \(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right).x=2012.\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\)
=> x = 2012