a) Ta có: \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}=1\text{-}\frac{1}{3}+\frac{1}{3}\text{-}\frac{1}{5}+...+\frac{1}{11}\text{-}\frac{1}{13}=1\text{-}\frac{1}{13}=\frac{12}{13}\)

Thay vào ta có:

\(\frac{12}{13}+x=\frac{24}{13}\Rightarrow x=\frac{24}{13}\text{-}\frac{12}{13}\Rightarrow x=\frac{12}{13}\)

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

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

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

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

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

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

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

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

Bài 1:

\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{101}\right|=101x\)

Ta thấy:

\(VT\ge0\Rightarrow VP\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)

\(\Rightarrow\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+...+\left(x+\frac{1}{101}\right)=101x\)

\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{101}\right)=0\)

\(\Rightarrow10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)=0\)

\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)=0\)

\(\Rightarrow10x+\left(1-\frac{1}{11}\right)=0\)

\(\Rightarrow10x+\frac{10}{11}=0\)

\(\Rightarrow10x=-\frac{10}{11}\Rightarrow x=-\frac{1}{11}\)(loại,vì x\(\ge\)0)

Bài 2:

Ta thấy: \(\begin{cases}\left(2x+1\right)^{2008}\ge0\\\left(y-\frac{2}{5}\right)^{2008}\ge0\\\left|x+y+z\right|\ge0\end{cases}\)

\(\Rightarrow\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|\ge0\)

Mà \(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)

\(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)

\(\Rightarrow\begin{cases}\left(2x+1\right)^{2008}=0\\\left(y-\frac{2}{5}\right)^{2008}=0\\\left|x+y+z\right|=0\end{cases}\)\(\Rightarrow\begin{cases}2x+1=0\\y-\frac{2}{5}=0\\x+y+z=0\end{cases}\)

\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\x+y+z=0\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{2}+\frac{2}{5}+z=0\end{cases}\)

\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{10}=-z\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{1}{10}\end{cases}\)

Bài 3:

a)\(2009-\left|x-2009\right|=x\)

\(\Rightarrow\left|x-2009\right|=2009-x\)

\(\Rightarrow\left|x-2009\right|=-\left(x-2009\right)\)

Vì GTTĐ của số âm bằng số đối của nó

\(\Rightarrow x-2009\le0\)

\(\Rightarrow x\le2009\)

Vậy với mọi \(x\le2009\) đều thỏa mãn

b)\(\left|3x+2\right|=\left|5x-3\right|\)

\(\Rightarrow3x+2=5x-3\) hoặc \(3x+2=3-5x\)

\(\Rightarrow2x=5\) hoặc \(8x=1\)

\(\Rightarrow x=\frac{5}{2}\) hoặc \(x=\frac{1}{8}\)

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

mình chưa tìm ra câu trả lời xin lỗi

Anh chỉ giải câu a thôi, câu b anh thấy nó bình thường mà.

Cộng vào mỗi phân số thêm 1 đơn vị được:

\(\frac{x+2013}{2009}+\frac{x+2013}{2010}=\frac{x+2013}{2011}+\frac{x+2013}{2012}\).

Tới đây tự làm tiếp nhá.

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