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 \(x=\frac{1}{8}\)

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)

\(\Rightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)

\(\Rightarrow2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)

\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)

\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{2009}\div2\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{4018}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2007}{4018}\)

\(\Rightarrow\frac{1}{x+1}=\frac{2}{4018}=\frac{1}{2009}\)

\(\Rightarrow x+1=2009\)

\(\Rightarrow x=2008\)

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)

=>\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2007}{4018}\)(nhân cả hai vế với \(\frac{1}{2}\))

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\)= \(\frac{2007}{4018}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2007}{4018}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{4018}\)

\(\frac{1}{x+1}\)=\(\frac{1}{2}-\frac{2007}{4018}\)

\(\frac{1}{x+1}=\frac{1}{2009}\)

x+1=2009

x=2009-1=2008

Vậy x bằng 2008

x=2008

Ta có : \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x.\left(x+1\right)}=\frac{2008}{2009}\)

\(\Leftrightarrow\)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}=\frac{2008}{2009}\)

\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2008}{2009}\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{2008}{2009}\)

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

\(\Leftrightarrow x+1=2009\)

\(\Leftrightarrow x=2008\)

Vậy x = 2008

\(=>\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{x.\left(x+1\right)}=\frac{2008}{2009}\)

\(=>1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2008}{2009}\)

\(=>1-\frac{1}{x+1}=\frac{2008}{2009}\)

\(=>\frac{x}{x+1}=\frac{2008}{2009}=>x=2008\)

Thế :

cái cuối đau cần đâu 

Giải toán trên mạng - Giúp tôi giải toán - Hỏi đáp, thảo luận về toán học - Học toán với OnlineMath

Em tham khảo nhé!

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)

=>\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)

=> \(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)

=> \(2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)

=> \(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)

=> \(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)

=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{2009}:2=\frac{2007}{4018}\)

=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{2007}{4018}=\frac{2009}{4018}-\frac{2007}{4018}\)

=> \(\frac{1}{x+1}=\frac{2}{4018}=\frac{1}{2009}\)

=> \(1\cdot2009=1\left(x+1\right)\)

=> \(x+1=2009\Rightarrow x=2009-1=2008\)

Vậy x = 2008

Chúc bn hk tốt !

\(\hept{\begin{cases}\left|x+\frac{1}{2009}\right|\ge0\\....\\\left|x+\frac{2008}{2009}\right|\ge0\end{cases}\Rightarrow\left|x+\frac{1}{2009}\right|+\left|x+\frac{2}{2009}\right|+....\left|x+\frac{2008}{2009}\right|\ge0}\)

\(\Rightarrow2009x\ge0\Rightarrow x\ge0\)

\(\Rightarrow\hept{\begin{cases}\left|x+\frac{1}{2009}\right|=x+\frac{1}{2009}\\....\\\left|x+\frac{2008}{2009}\right|=x+\frac{2008}{2009}\end{cases}\Rightarrow x+\frac{1}{2009}+...+x+\frac{2008}{2009}}=2009x\)

\(2008x+201840=2009x\Rightarrow x=201840\)

p/s: cách làm thì khá ok, nhưng kq không chắc lắm nhé, có gì bn tính lại nha

Boul đẹp trai_tán gái đổ 100% sai 100%

Sao dòng cuối lại tek ? Các phân số ấy cộng vào không thể là 201840

Về hướng làm thì đúng nhưng chỉ đúng đến bước phá trị thôi 

Tham khảo cách làm nhưg nhớ đổi đoạn cuối nhé !

a sorry cộng lại quên mẹ chia cho 2009 :> mà tính máy tính ko hiểu sao cx sai lun, đổi lại kq nha :>

\(x=\frac{2017036}{2009}=1004\)

\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1000}{2002}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1000}{2002}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1000}{2002}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{1000}{2002}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2002}\)

<=>x+1=2002

=>x=2001

dap an :

X=X