1/1.2+1/2.3+.....+1/x.(x+1)=2008/2009

=>1/1-1/2+1/2-1/3+.....+1/x-1/x+1=2008/2009

=>1/1+(-1/2+1/2)+(-1/3+1/3)+....+(-1/x+1/x)-1/x+1=2008/2009

=>1/1+0+0+.....+0-1/x+1=2008/2009

=>1-1/x+1=2008/2009

=>1/x+1=1-2008/2009=1/2009

=>x+1=2009

=>x=2008

 vậy x=2008

có cần cách làm k

x là nhân ak? ôi zời

\(a)\) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)

\(\Leftrightarrow\)\(\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=-4+4\)

\(\Leftrightarrow\)\(\frac{x+1+99}{99}+\frac{x+2+98}{98}+\frac{x+3+97}{97}+\frac{x+4+96}{96}=0\)

\(\Leftrightarrow\)\(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)

\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)

Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\)

Nên \(x+100=0\)

\(\Rightarrow\)\(x=-100\)

Vậy \(x=-100\)

Chúc bạn học tốt ~ 

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

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

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

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

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

\(\Leftrightarrow\)\(x+1=2009\)

\(\Leftrightarrow\)\(x=2009-1\)

\(\Leftrightarrow\)\(x=2008\)

Vậy \(x=2008\)

Chúc bạn học tốt ~ 

a) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)

\(\Leftrightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\)\(\left(\frac{x+4}{96}+1\right)=-4+4\)

\(\Leftrightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)

Mà  \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\)

\(\Rightarrow x+100=0\)

\(\Leftrightarrow x=-100\)

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

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

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

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

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 :>

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

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

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{499}{500}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{499}{500}\)

\(1-\frac{1}{x+1}=\frac{499}{500}\)

\(\frac{1}{x+1}=1-\frac{499}{500}=\frac{1}{500}\)

=> x + 1 = 500

=> x = 500 - 1

=> x = 499

Vậy x = 499

1/1.2 + 1/2.3 + 1/3.4 +...+ 1/x.(x+1)=499/500

1 - 1/2 + 1/2 -1/3 + 1/3 - 1/4 +...+ 1/x -1/(x+1) =499/500

1-1/(x+1)=499/500

=>x/(x+1)=499/500

=>x=499

Sửa đề:

\(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{x\times\left(x+1\right)}=\frac{9}{10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{9}{10}\)

\(1-\frac{1}{x}=\frac{9}{10}\)

\(\frac{1}{x}=1-\frac{9}{10}=\frac{1}{10}\)

Vậy, x = 10.

Ko bt có right ko?

Nhầm.

Chuyển \(1-\frac{1}{x}\)thành \(1-\frac{1}{x+1}\)

\(1-\frac{1}{x+1}=\frac{9}{10}\)

\(\frac{1}{x+1}=1-\frac{9}{10}=\frac{1}{10}\)

Vậy x = 10 - 1 = 9

Thế ms right chứ!

\(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{x\left(x+1\right)}=\frac{9}{10}\)

\(\Leftrightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{9}{10}\)

\(\Leftrightarrow\frac{1}{1}-\frac{1}{x+1}=\frac{9}{10}\Leftrightarrow\frac{1}{x+1}=\frac{1}{1}-\frac{9}{10}=\frac{1}{10}\)

\(\Leftrightarrow x+1=10\Rightarrow x=10-1=9\)

Vậy x = 9 

Ta có: 1/1x2 + 1/2x3 + 1/3x4 +...+ 1/X x (X + 1) = 499/500

=> 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +...+ 1/X - 1/(X + 1) = 499/500

=> 1 - 1/(X + 1) = 499/500

=>      1/(X + 1) = 1 - 499/500

=>      1/(X + 1) = 1/500

=>          X + 1 = 500

=>          X       = 500 - 1

=>          X       = 499 

Đáp số: X = 499