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

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

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

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

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

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2008}{2009 }\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2008}{2009}\)
\(1-\frac{1}{x+1}=\frac{2008}{2009}\)
\(\frac{x+1-1}{x+1}=\frac{2008}{2009}\)
\(\frac{x}{x+1}=\frac{2008}{2009}\)
\(2009x=2008\left(x+1\right)\)
\(2009x=2008x+2008\)
\(2009x-2008x=2008\)
\(x=2008\)
Vậy x=2008

Ta có

1/x.(x+1) =2008-1/1.2-1/2.3-....

tự làm nhé!!

$\textbf{a)}$

$\dfrac{x-1}{2}+\dfrac{x-2}{5}=\dfrac14+\dfrac{x-7}{10}$

$\Leftrightarrow\dfrac{5(x-1)+2(x-2)}{10}=\dfrac14+\dfrac{x-7}{10}$

$\Leftrightarrow\dfrac{7x-9}{10}=\dfrac14+\dfrac{x-7}{10}$

$\Leftrightarrow14x-18=5+2x-14$

$\Leftrightarrow12x=9$

$\Leftrightarrow x=\dfrac34.$

$\textbf{b)}$

$\dfrac{3-2}{2x-3}=\dfrac25+\dfrac1{2x-3}-\dfrac32$

$\Leftrightarrow1-\dfrac2{2x-3}=-\dfrac{11}{10}+\dfrac1{2x-3}$

$\Leftrightarrow\dfrac{21}{10}=\dfrac3{2x-3}$

$\Leftrightarrow21(2x-3)=30$

$\Leftrightarrow42x=93$

$\Leftrightarrow x=\dfrac{31}{14}.$

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

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

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

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

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 !

câu hỏi hay......nhưng tui xin nhường cho các bn khác

Hãy tích đúng cho tui nha

THANKS

a. 2006/2005 x 2007/2006 x 2008/2007 x 2009/2008 x 2010/2009'

= 2006 x 2007 x 2008 x 2009 x 2010 / 2005 x 2006 x 2007 x 2008 x 2009

= 2010/2005

= 402/401

\(\left(1+\frac{1}{2005}\right)x\left(1+\frac{1}{2006}\right)x\left(1+\frac{1}{2007}\right)x\left(1+\frac{1}{2008}\right)x\left(1+\frac{1}{2009}\right)\)

\(=\frac{2006}{2005}x\frac{2007}{2006}x\frac{2008}{2007}x\frac{2009}{2008}x\frac{2010}{2009}\)

\(=\frac{2010}{2005}\)

\(=\frac{402}{401}\)

a)\(\frac{5}{2}-3\left(\frac{1}{3}-x\right)=\frac{1}{4}-7x\)

\(\Leftrightarrow\frac{5}{2}-1+x=\frac{1}{4}-7x\)

\(\Leftrightarrow8x=-\frac{5}{4}\)

\(\Leftrightarrow x=-\frac{5}{32}\)

c)\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)

\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)

\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2003}\)

\(\Leftrightarrow x+1=2003\)

\(\Leftrightarrow x=2002\)