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

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

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

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

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

Nguyễn Khánh Linh

a,

\(\left[1+\frac{1}{2005}\right].\left[1+\frac{1}{2006}\right].\left[1+\frac{1}{2007}\right].\left[1+\frac{1}{2008}\right].\left[1+\frac{1}{2009}\right]\)

  \(\Rightarrow\left[\frac{2005}{2005}+\frac{1}{2005}\right]\left[\frac{2006}{2006}+\frac{1}{2006}\right]\left[\frac{2007}{2007}+\frac{1}{2007}\right]\)  \(\left[\frac{2008}{2008}+\frac{1}{2008}\right]\left[\frac{2009}{2009}+\frac{1}{2009}\right]\)                                                                                                                                                               

\(\Rightarrow\frac{2006}{2005}.\frac{2007}{2006}.\frac{2008}{2007}.\frac{2009}{2008}.\frac{2010}{2009}\)

\(\Rightarrow\frac{2010}{2005}=\frac{402}{401}\)

tick cho mình rồi mình làm cho

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

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

1. a) \(\frac{3}{4}-\frac{-1}{2}+\frac{1}{3}=\frac{3}{4}+\frac{1}{2}+\frac{1}{3}=\frac{9}{12}+\frac{6}{12}+\frac{4}{12}=\frac{19}{12}\)

   b) \(5\frac{5}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}\)

\(=\frac{140}{27}-\frac{5}{27}+\frac{7}{23}+\frac{16}{23}+\frac{1}{2}\)

\(=\frac{135}{27}+\frac{23}{23}+\frac{1}{2}\)

\(=5+1+0,5=6,5\)

2) a) 1/2 + 2/3x = 1/4

=> 2/3x            = 1/4 - 1/2

=> 2/3x            = -1/4

=> x                = -1/4 : 2/3

=> x                = -3/8

b) 3/5 + 2/5 : x = 3 1/2

=> 3/5 + 2/5 : x = 7/2

=>         2/5 : x  = 7/2 - 3/5

=>         2/5 : x  = 29/10

=>               x    = 2/5 : 29/10

=>               x    = 4/29

c) x+4/2004 + x+3/2005 = x+2/2006 + x+1/2007

=> x+4/2004 + 1 + x+3/2005 + 1 = x+2/2006 + 1 + x+1/2007 + 1

=>   x+2008/2004 + x+2008/2005 = x+2008/2006 + x+2008/2007

=>  x+2008/2004 + x+2008/2005 - x+2008/2006 - x+2008/2007 = 0

=> (x+2008). (1/2004 + 1/2005 - 1/2006 - 1/2007) = 0

Vì 1/2004 + 1/2005 - 1/2006 - 1/2007 khác 0

Nên x + 2008 = 0 <=> x = -2008

Vậy x = -2008

1,a,\(\frac{3}{4}-\frac{-1}{2}+\frac{1}{3}=\frac{3}{4}+\frac{2}{4}+\frac{1}{3}=\frac{5}{4}+\frac{1}{3}=\frac{15}{12}+\frac{4}{12}=\frac{19}{12}\)

  b, \(5\frac{5}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}=\frac{140}{27}-\frac{5}{27}+\frac{7}{23}+\frac{16}{23}+\frac{1}{2}=\frac{135}{27}+\frac{23}{23}+\frac{1}{2}=5+1+\frac{1}{2}=\frac{13}{2}\)2,a,\(\frac{1}{2}+\frac{2}{3}.x=\frac{1}{4}\)

    <=>\(\frac{2}{3}.x=-\frac{1}{2}\)

   <=>\(x=-\frac{3}{4}\)

b,\(\frac{3}{5}+\frac{2}{5}\div x=3\frac{1}{2}\)

 <=>\(\frac{2}{5x}=\frac{29}{10}\)

 <=>\(x=\frac{29}{4}\)

c,\(\frac{x+4}{2004}+\frac{x+3}{2005}=\frac{x+2}{2006}+\frac{x+1}{2007}\)

<=> \(\frac{x+4}{2004}+1+\frac{x+3}{2005}+1=\frac{x+2}{2006}+1+\frac{x+1}{2007}+1\)

<=>\(\frac{x+2008}{2004}+\frac{x+2008}{2005}=\frac{x+2008}{2006}+\frac{x+2008}{2007}\)

<=>\(\left(x+2008\right)\left(\frac{1}{2004}+\frac{1}{2005}-\frac{1}{2006}-\frac{1}{2007}\right)\)=0

<=>x+2008=0 vì cái ngoặc còn lại\(\ne0\)

<=>x=-2008

 Vậy x=-2008

Bạn nhớ tk cho mình vì mình đã chăm chỉ làm hết bài bạn hỏi nha!

1.

a) \(\frac{3}{4}-\frac{-1}{2}+\frac{1}{3}=\frac{3}{4}+\frac{1}{2}+\frac{1}{3}=\frac{19}{12}\)

b) \(5\frac{5}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}\)

\(=\frac{140}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}\)

\(=5+1+\frac{1}{2}=\frac{13}{2}\)

2.

a) \(\frac{1}{2}+\frac{2}{3}x=\frac{1}{4}\)

\(\frac{2}{3}x=-\frac{1}{4}\)

\(x=-\frac{3}{8}\)

b) \(\frac{3}{5}+\frac{2}{5}:x=3\frac{1}{2}\)

\(\frac{2}{5}x=\frac{7}{2}-\frac{3}{5}\)

\(\frac{2}{5}x=\frac{29}{10}\)

\(x=\frac{29}{4}\)

c) \(\frac{x+4}{2004}+\frac{x+3}{2005}=\frac{x+2}{2006}+\frac{x+1}{2007}\)

\(\left(\frac{x+4}{2004}+1\right)+\left(\frac{x+3}{2005}+1\right)=\left(\frac{x+2}{2006}+1\right)\left(\frac{x+1}{2007}+1\right)\)

\(\frac{x+2008}{2004}+\frac{x+2008}{2005}-\frac{x+2008}{2006}-\frac{x+2008}{2007}=0\)

\(\left(x+2008\right)\left(\frac{1}{2004}+\frac{1}{2005}+\frac{1}{2006}+\frac{1}{2007}\right)=0\)

\(x+2008=0\)

\(x=-2008\) ( vì \(\frac{1}{2004}+\frac{1}{2005}+\frac{1}{2006}+\frac{1}{2007}\ne0\))

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

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

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

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

\(=>x+2010=0\)

\(=>x=-2010\)

\(pt\Leftrightarrow\frac{x}{2009}+\frac{1}{2009}+\frac{x}{2008}+\frac{2}{2008}=\frac{x}{3}+\frac{2007}{3}+\frac{x}{4}+\frac{2006}{4}\Leftrightarrow\frac{x}{2009}+\frac{x}{2008}-\frac{x}{3}-\frac{x}{4}=\frac{2006}{4}+\frac{2007}{3}-\frac{1}{1008}-\frac{1}{2009}\Leftrightarrow x\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{3}-\frac{1}{4}\right)=\frac{2006}{4}+\frac{2007}{3}-\frac{1}{1008}-\frac{1}{2009}\Leftrightarrow x=\frac{\frac{2006}{4}+\frac{2007}{3}-\frac{1}{1008}-\frac{1}{2009}}{\frac{1}{2009}+\frac{1}{2008}-\frac{1}{3}-\frac{1}{4}}=-2010\)

2010 nha

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

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

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

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

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

Vì \(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\ne0\)

=>x+2010=0

=>x=-2010

Vậy x = -2010

chuanhieu cho lam