\(x\)là dấu nhân hả bạn? Nếu vậy thì mk làm cho nhé

\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot....\cdot\left(1-\frac{1}{20}\right)\)

\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot.......\cdot\frac{17}{18}\cdot\frac{18}{19}\cdot\frac{19}{20}=\frac{1}{20}\)

Vậy \(A=\frac{1}{20}\)

\(B=1\frac{1}{2}\cdot1\frac{1}{3}\cdot1\frac{1}{4}\cdot........\cdot1\frac{1}{2005}\cdot1\frac{1}{2006}\cdot1\frac{1}{2007}\)

\(B=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot......\cdot\frac{2006}{2005}\cdot\frac{2007}{2006}\cdot\frac{2008}{2007}=\frac{2008}{2}=1004\)

Vậy \(B=1004\)

DẤU CHẤM LÀ DẤU NHÂN

a, 

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{19}{20}=\frac{1}{20}\)

b, \(1\frac{1}{2}.1\frac{1}{3}....1\frac{1}{2017}=\frac{3}{2}.\frac{4}{3}....\frac{2018}{2017}=\frac{2018}{2}=1009\)

a) \(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)

<=> \(\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)-\left(\frac{x-3}{2007}-1\right)-\left(\frac{x-4}{2006}-1\right)=0\)

<=> \(\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)

<=> \(\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)

<=> x - 2010 = 0 Vì \(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\ne0\)

<=> x = 2010

\(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|=4\left(x-4\right)\)

Ta thấy : \(\left|x-1\right|\ge0;\left|x-2\right|\ge0;\left|x-3\right|\ge0\)

=> \(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|\ge0\)

=> 4 ( x - 4 ) \(\ge0\). Mà 4 > 0 => \(x-4\ge0=>x\ge4\)hay

\(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|=4\left(x-4\right)=>x-1+x-2+x-3=4\left(x-4\right)\) => 3x - 6 = 4x - 16

=> -6+16 = 4x - 3x => x = 10

a)  \(\Leftrightarrow\frac{x+7}{2003}+1+\frac{x+4}{2006}+1-\frac{x-1}{2011}-1-\frac{x-5}{2015}-1=0\)

     \(\Leftrightarrow\frac{x+2010}{2003}+\frac{x+2010}{2006}-\frac{x+2010}{2011}-\frac{x+2010}{2015}=0\)

     \(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2003}+\frac{1}{2006}-\frac{1}{2011}-\frac{1}{2015}\right)=0\)

     \(\Leftrightarrow x+2010=0\) ( vì 1/2003  +  1/2006  --  1/2011  -- 1/2015   \(\ne\)0)

    \(\Leftrightarrow x=-2010\)

câu b làm tương tự (có gì không hiểu hỏi mk nha) >v<

c) \(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)

\(\Leftrightarrow\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)=\left(\frac{x-3}{2007}-1\right)+\left(\frac{x-4}{2006}-1\right)\)

\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)

\(\Leftrightarrow\left(x-2010\right).\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)

\(\Leftrightarrow x-2010=0\)

\(\Leftrightarrow x=0+2010\)

\(\Rightarrow x=2010\)

Vậy \(x=2010.\)

Mình chỉ làm câu c) thôi nhé.

Chúc bạn học tốt!

1)   \(\frac{x+4}{2005}\)\(+\)\(\frac{x+3}{2006}\)= \(\frac{x+2}{2007}\)\(+\)\(\frac{x+1}{2008}\)

\(\Leftrightarrow\)   \(\frac{x+4}{2005}\)\(+\)1 \(+\)\(\frac{x+3}{2006}\)\(+\)1 = \(\frac{x+2}{2007}\)\(+\)1 \(+\)\(\frac{x+1}{2008}\)\(+\)1

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

\(\Leftrightarrow\)(x + 2009)(1/2005 + 1/2006) = (x + 2009)(1/2007 + 1/2008)

\(\Leftrightarrow\)(x + 2009)(1/2005 + 1/2006 - 1/2007 - 1/2008) = 0

Ta thấy:  1/2005 + 1/2006 - 1/2007 - 1/2008 \(\ne\)0

\(\Leftrightarrow\)x + 2009 = 0

\(\Leftrightarrow\)x = -2009

\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=\left(x-1\right)\left(x-2\right)x=0\)

tìm đc x=0;1;2

\(\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{8}\)

\(\frac{8\left(x+2007\right)}{8\left(x+2000\right)\left(x+2007\right)}-\frac{8\left(x+2000\right)}{8\left(x+2000\right)\left(x+2007\right)}=\frac{7\left(x+2000\right)\left(x+2007\right)}{8\left(x+2000\right)\left(x+2007\right)}\)

\(8x+8.2007-8x+8.2000=7\left(x^2+4007x+2000.2007\right)\)

\(8.7-7\left(x^2+4007x+2000.2007\right)=0\)

\(7\left(8-x^2-4007x-2000.2007\right)=0\)

\(8-x^2-4007x-2000.2007=0\)

\(x^2+4007x+4013992=0\)

\(\left(x^2+2008x\right)+\left(1999x+4013992\right)=0\)

\(\left(x+2008\right)\left(x+1999\right)=0\)

\(\hept{\begin{cases}x=-2008\\x=-1999\end{cases}}\)

\(\frac{1}{\left(x+2000\right)\left(x+2001\right)}+\frac{1}{\left(x+2001\right)\left(x+2002\right)}+\frac{1}{\left(x+2006\right)\left(x+2007\right)}=\frac{7}{8}\)

\(\frac{1}{x+2000}-\frac{1}{x+2001}+\frac{1}{x+2001}-\frac{1}{x+2002}+...+\frac{1}{x+2006}-\frac{1}{x+2007}=\frac{7}{8}\)

\(\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{8}\)

phần đầu mk thiếu điều kiện,bn tự bổ sung nha

Bài 1 :\(a,=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{100^2}{99.101}\)

           \(=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4...101}\)

          \(=100.\frac{2}{101}=\frac{200}{101}\)

\(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

=> \(\frac{1}{x+2000}-\frac{1}{x+2001}+\frac{1}{x+2001}-\frac{1}{x+2002}+....+\frac{1}{x+2006}-\frac{1}{x+2007}=\frac{7}{8}\)

<=> \(\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{8}\)

<=> \(\frac{7}{\left(x+2000\right)\left(x+2007\right)}=\frac{7}{8}\Leftrightarrow\left(x+2000\right)\left(x+2007\right)=8\)

=> x = -1999 hoặc x = - 2008

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