Tìm y : 1/3 + 1/6 + 1/10 +...+ 1/Yx(Y +1 : 2 ) = 2009/2011
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\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)
\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)
\(\frac{1}{x+1}=\frac{1}{2011}\)
\(x+1=2011\)
\(x=2010\)
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Nhân cả 2 vế với 1/2 ta có:
1/6+ 1/12+ 1/20+... +1/x(x+1)=2009/4022
1/2.3+ 1/3.4+ 1/4.5+... +1/x(x+1)=2009/4022
1/2- 1/3+ 1/3-1/4+ 1/4- 1/5+...+1/x- 1/x+1=2009/4022
1/2- 1/x+1=2009/4022
x+1/2.(x+1)- 2/2.(x+1)=2009/4022
x+1-2/2.(x+1)=2009/4022
x-1/2.(x+1)=2009/4022
=>(x-1). 4022= 2009.2.(x+1)
4022.x- 4022= 4018.x+ 4018
4022.x-4018.x= 4018+ 4022
x.(4022- 4018)= 8040
x.4 = 8040
x = 8040: 4
x =2010
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1.1/3+1/6+1/10+...+2/x.(x+1)=2007/2009
=>2/6+2/12+2/20+...+2/x.(x+1)=2007/2009
=>1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/(x+1)=2007/2009:2
=>1/2-1/(x+1)=2007/4018
=>1/(x+1)=1/2-2007/4018
=>1/x+1=1/2009
=>x+1=2009
=>x=2009-2008
=>x=1
vậy x=1