Bài 1:

Ta có: \(\frac{497}{-499}=-\frac{497}{499}>-\frac{499}{499}=-1\left(1\right)\)

\(-\frac{2345}{2341}< -\frac{2341}{2341}=-1\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\frac{497}{-499}>-\frac{2345}{2341}\)

Bài 2:

\(\frac{x+5}{2005}+\frac{x+6}{2004}=\frac{x+7}{2003}+3=0\)

\(\Rightarrow\frac{x+5}{2005}+\frac{x+6}{2004}+\frac{x+7}{2003}+3=0\)

\(\Rightarrow\frac{x+5}{2005}+1+\frac{x+6}{2004}+1+\frac{x+7}{2003}+1=0\)

\(\Rightarrow\frac{x+2010}{2005}+\frac{x+2010}{2004}+\frac{x+2010}{2003}=0\)

\(\Rightarrow\left(x+2010\right)\times\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)=0\)

Vì \(\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)\ne0\Rightarrow x+2010=0\)

\(\Rightarrow x=0-2010=-2010\)

Vậy x = -2010

a) 5x.(x+3/4) = 0

=> x = 0

x+3/4 = 0 => x = -3/4

b) \(\frac{x+7}{2010}+\frac{x+6}{2011}=\frac{x+5}{2012}+\frac{x+4}{2013}.\)

\(\Rightarrow\frac{x+7}{2010}+\frac{x+6}{2011}-\frac{x+5}{2012}-\frac{x+4}{2013}=0\)

\(\frac{x+7}{2010}+1+\frac{x+6}{2011}+1-\frac{x+5}{2012}-1-\frac{x+4}{2013}-1=0\)

\(\left(\frac{x+7}{2010}+1\right)+\left(\frac{x+6}{2011}+1\right)-\left(\frac{x+5}{2012}+1\right)-\left(\frac{x+4}{2013}+1\right)=0\)

\(\frac{x+2017}{2010}+\frac{x+2017}{2011}-\frac{x+2017}{2012}-\frac{x+2017}{2013}=0\)

\(\left(x+2017\right).\left(\frac{1}{2010}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}\right)=0\)

=> x + 2017 = 0

x = -2017

a) để 2x - 3 > 0

=> 2x > 3

x > 3/2

b) 13-5x < 0

=> 5x < 13

x < 13/5

c) \(\frac{x+3}{2x-1}>0\)

=> x + 3 > 0

x > -3

d) \(\frac{x+7}{x+3}=\frac{x+3+4}{x+3}=1+\frac{4}{x+3}\)

Để x+7/x+3 < 1

=> 1 + 4/x+3 < 1

=> 4/x+3 < 0

=> không tìm được x thỏa mãn điều kiện

\(x-\left(\frac{5}{6}-x\right)=x-\frac{2}{3}\)

\(x-\frac{5}{6}+x-x+\frac{2}{3}=0\)

\(x-\frac{1}{6}=0\)

\(x=\frac{1}{6}\)

vậy \(x=\frac{1}{6}\)

=> x- (x-2/3)=5/6-x

=> x-x+2/3=5/6-x

=> 2/3 = 5/6-x

=> x= 5/6-2/3 = 1/6

              \(x-\left(\frac{5}{6}-x\right)=x-\frac{2}{3}\)\(x-\left(\frac{5}{6}-x\right)=x-\frac{2}{3}\)

              \(\Rightarrow x-\frac{5}{6}+x=x-\frac{2}{3}\)

               \(\Rightarrow2x-\frac{5}{6}=x-\frac{2}{3}\)

               \(\Rightarrow2x-x=\frac{5}{6}-\frac{2}{3}\)

                \(\Rightarrow x=\frac{1}{6}\)

                                   Vậy \(x=\frac{1}{6}\)

Bài 1:

a: =>7x-21=2x+2

=>5x=23

=>x=23/5

b: =>(x+5)^2=100

=>x+5=10 hoặc x+5=-10

=>x=-15 hoặc x=5

Bai 2: 

a: 2x-3>0

=>2x>3

=>x>3/2

b: =>13-5x<0

=>5x>13

=>x>13/5

c: =>2x-1>0 hoặc x+3<0

=>x>1/2 hoặc x<-3

d: =>(x+7-x-3)/(x+3)<0

=>x+3<0

=>x<-3

\(\frac{x+3}{2007}-\frac{x+3}{2008}=\frac{x+3}{2010}-\frac{x+3}{2009}\)

\(\Rightarrow\frac{x+3}{2007}-\frac{x+3}{2008}-\frac{x+3}{2010}+\frac{x+3}{2009}=0\)

\(\left(x+3\right)\cdot\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2010}+\frac{1}{2009}\right)=0\)

mà \(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2010}+\frac{1}{2009}\ne0\)

=> x + 3 = 0

x = -3