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2.[1/6+1/12+1/20+...+1/x.(x+1)]=2009/2011
2.[1/2.3+1/3.4+1/4.5+...+1/x(x+1)]=2009/2011
1/2-1/3+1/3-1/4+...+1/x-1/(x+1)=2009/4022
1/2-1/(x+1)=2009/4022
1/(x+1)=1/2001
x+1=2011
x=2010
\(=>\frac{2}{3.2}+\frac{2}{6.2}+\frac{2}{10.2}+...+\frac{2}{x.\left(x+1\right):2.2}=\frac{2009}{2011}\)
\(=>\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x.\left(x+1\right)}=\frac{2009}{2011}\)
\(=>2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2009}{2011}\)
\(=>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{2009}{2011}:2\)
\(=>1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{4022}\)
\(=>1-\frac{1}{x+1}=\frac{2009}{4022}\)
\(=>\frac{1}{x+1}=1-\frac{2009}{4022}\)
\(=>\frac{1}{x+1}=\frac{2013}{4022}\)
\(=>\frac{2013}{2013.\left(x+1\right)}=\frac{2013}{4022}\)
=>2013.(x+1)=4022
=>x+1=4022/2013
=>x=4022/2013-1
=>x=2009/2013
có \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right):2}=\frac{2009}{2011}\)
tách vế trái đặt là A
ta lại có\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right):2}\)
\(\frac{1}{2}A=\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{x.\left(x+1\right):2}\right)\)
\(\frac{1}{2}A=\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x.\left(x+1\right)}\)
\(\frac{1}{2}A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}\)
\(\frac{1}{2}A=\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{x}-\frac{1}{x+1}\right)\)
\(\frac{1}{2}A=\frac{1}{2}-\frac{1}{x+1}\)
\(A=\left(\frac{1}{2}-\frac{1}{x+1}\right):\frac{1}{2}\)
\(A=1+\frac{1}{\left(x+1\right):2}\)
ta thế vào vế trái vào vế phải
ta có\(1+\frac{1}{\left(x+1\right):2}=\frac{2009}{2011}\)
\(\frac{1}{\left(x+1\right):2}=\frac{2009}{2011}-1\)
\(\frac{1}{\left(x+1\right):2}=\frac{2009}{2011}-\frac{2011}{2011}=-\frac{2}{2011}\)
\(-\frac{2}{-\left(x+1\right)}=-\frac{2}{2011}\)
thấy hai tử bằng nhau
\(\Rightarrow-\left(x+1\right)=2011\)
\(\Rightarrow\left(x+1\right)=-2011\)
\(\Rightarrow x=-2011-1=-2012\)
1+1/3+1/6+1/10+...+1/x.(x+1):2=1+2009/2011
=>2/6+2/12+2/20+...+2/x.(x+1)=2009/2011
=>2.(1/2.3+1/3.4+1/4.5+...+1/x.(x+1))=2009/2011
=>1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1=2009/2011:2
=>1/2-1/x+1=2009/4022
=>1/x+1=1/2-2009/4022
=>1/x+1=1/2001
=.x+1=2001
=>x=2001-1
=>x=2000
vậy x=2000
\(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{x\left(x+1\right):2}=\frac{2009}{2011}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{2011}:2\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2011}\)
\(\Leftrightarrow x+1=2011\)
\(\Leftrightarrow x=2010\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.......+\frac{1}{x\times\left(x+1\right)\div2}=\frac{2009}{2011}\)
\(2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+.......+\frac{1}{x\times\left(x+1\right)}\right)=\frac{2009}{2011}\)
\(2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)
\(2\times\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)
\(1-\frac{2}{x+1}=\frac{2009}{2011}\)
\(\frac{2}{x+1}=1-\frac{2009}{2011}\)
\(\frac{2}{x+1}=\frac{2}{2011}\)
\(x+1=2011\)
\(x=2011-1\)
\(\Rightarrow x=2010\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right):2}=\frac{2009}{2011}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}=\frac{2009}{4022}\)(nhân mỗi vế với 1/2)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{2009}{4022}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{4022}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2009}{4022}=\frac{1}{2011}\)
\(\Rightarrow x+1=2011\Rightarrow x=2010\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=\frac{2009}{2011}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}\right)=\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\)\(=\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\)\(=\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2011}\)
\(\Rightarrow x+1=2011\)
\(\Rightarrow x=2010\)
1) Tính :
a) \(\left(2008.2009.2010.2011\right).\left(1+\frac{1}{2}:\frac{2}{3}-\frac{4}{3}\right)\)
\(=\left(2008.2009.2010.2011\right).\left(1+\frac{1}{3}-\frac{4}{3}\right)\)
\(=\left(2008.2009.2010.2011\right).\left(\frac{4}{3}-\frac{4}{3}\right)\)
\(=\left(2008.2009.2010.2011\right).0\)
\(=0\)
2) Tìm x
a) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{2013}:2\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2013}\)
\(\Rightarrow x+1=2013\)
\(\Rightarrow x=2012\)
b) \(\frac{1}{2}.\frac{1}{3}.\frac{1}{4}.\frac{1}{5}.\frac{1}{6}.\left(x-1,010\right)=\frac{1}{360}-\frac{1}{720}\)
\(\Rightarrow\frac{1}{2.3.4.5.6}.\left(x-1,01\right)=\frac{1}{720}\)
\(\Rightarrow\frac{1}{720}.\left(x-1,01\right)=\frac{1}{720}\)
\(\Rightarrow x-1,01=\frac{1}{720}:\frac{1}{720}\)
\(\Rightarrow x-1,01=1\)
\(\Rightarrow x=1+1,01\)
\(\Rightarrow x=2,01\)
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{\frac{x\left(x+2\right)}{2}}=1\frac{2009}{2011}\)
\(\Leftrightarrow1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{1}{x\left(x+2\right)}=1\frac{2009}{2011}\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{x\left(x+2\right)}=1\frac{2009}{2011}-1\)
\(\Leftrightarrow\left[2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)\right]=\frac{2009}{2011}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+2}\right)=\frac{2009}{2011}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+2}=\frac{2009}{2011}\div2=\frac{2009}{4022}\)
\(\Leftrightarrow\frac{1}{x+2}=\frac{1}{2}-\frac{2009}{4022}=\frac{1}{2011}\)
\(\Leftrightarrow x=2011-2=2009\)