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b, \(\frac{x+1}{2009}+\frac{x+2}{2009}=\frac{x+10}{2000}+\frac{x+11}{1999}\)
\(\Rightarrow\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)=\left(\frac{x+10}{2000}+1\right)+\left(\frac{x+11}{1999}+1\right)\)
\(\Rightarrow\frac{x+1+2009}{2009}+\frac{x+2+2008}{2008}=\frac{x+10+2000}{2000}+\frac{x+11+1999}{1999}\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}=\frac{x+2010}{2000}+\frac{x+2010}{1999}\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}-\frac{x+2010}{2000}-\frac{x+2010}{1999}=0\)
\(\Rightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2000}-\frac{1}{1999}\right)=0\)
Mà \(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2000}-\frac{1}{1999}\ne0\)
=> x + 2010 = 0 => x = -2010
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.(x+1)}=\frac{2007}{2009}\)
=> \(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2017}{2019}\)
=> \(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2017}{2019}\)
=> \(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
=> \(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{2009}:2\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{4018}\)
=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{2017}{4018}\)
=> \(\frac{1}{x+1}=\frac{1}{2019}\)
Vì 1 = 1
=> x + 1 = 2019
=> x = 2019 - 1
=> x = 2018
\(\frac{1985\times1987-1}{1984+1985\times1986}\)
\(=\frac{1985\times\left(1986+1\right)-1}{1984+1985\times1986}\)
\(=\frac{1985\times1986+1985-1}{1984+1985\times1986}\)
\(=\frac{1985\times1986+1984}{1984+1985\times1986}\)
\(=1\)
----------HOK TỐT-----------
Đây là cuộc thi nhé. cần sự công bằng. Mong em không tái phạm lần sau. Bạn sẽ bị khóa nick hoặc trừ 5000 điểm nhé!
BQT thân gửi em!
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