\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{\frac{x\left(x+2\right)}{2}}=1\frac{2009}{2011}\)

\(\Leftrightarrow1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{1}{x\left(x+2\right)}=1\frac{2009}{2011}\)

\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{x\left(x+2\right)}=1\frac{2009}{2011}-1\)

\(\Leftrightarrow\left[2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)\right]=\frac{2009}{2011}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+2}\right)=\frac{2009}{2011}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+2}=\frac{2009}{2011}\div2=\frac{2009}{4022}\)

\(\Leftrightarrow\frac{1}{x+2}=\frac{1}{2}-\frac{2009}{4022}=\frac{1}{2011}\)

\(\Leftrightarrow x=2011-2=2009\)

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

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

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

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

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

2010 nha bạn

khó quá đi bn à

Giả sử x là số nguyên dương

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)

\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)

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

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

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

\(\Leftrightarrow x+1=2011\)

\(\Rightarrow x=2010\)