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Nguyễn Vũ Minh Hiếu
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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

\(\Rightarrow-\left(x+1\right)=2011\)

\(\Rightarrow\left(x+1\right)=-2011\)

\(\Rightarrow x=-2011-1=-2012\)

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Nguyễn Thị Thanh Trang
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Lê Nguyễn Bảo Châu
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Lightning Farron
14 tháng 10 2016 lúc 11:07

\(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{x\left(x+1\right):2}=\frac{2009}{2011}\)

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)

\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)

\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{2011}:2\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2011}\)

\(\Leftrightarrow x+1=2011\)

\(\Leftrightarrow x=2010\)

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Trần Thị Bảo Trân
14 tháng 10 2016 lúc 11:15

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.......+\frac{1}{x\times\left(x+1\right)\div2}=\frac{2009}{2011}\)

\(2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+.......+\frac{1}{x\times\left(x+1\right)}\right)=\frac{2009}{2011}\)

\(2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)

\(2\times\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)

\(1-\frac{2}{x+1}=\frac{2009}{2011}\)

\(\frac{2}{x+1}=1-\frac{2009}{2011}\)

\(\frac{2}{x+1}=\frac{2}{2011}\)

\(x+1=2011\)

\(x=2011-1\)

\(\Rightarrow x=2010\)

 

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