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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
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
\(\frac{x+10}{2008}+\frac{x+9}{2009}=\frac{x+8}{2010}+\frac{x+7}{2011}\)
\(\left(\frac{x+10}{2008}+1\right)+\left(\frac{x+9}{2009}+1\right)=\left(\frac{x+8}{2010}+1\right)+\left(\frac{x+7}{2011}+1\right)\)
\(\frac{x+2018}{2008}+\frac{x+2018}{2009}=\frac{x+2018}{2010}+\frac{x+2018}{2011}\)
\(\frac{x+2018}{2008}+\frac{x+2018}{2009}-\frac{x+2018}{2010}-\frac{x+2018}{2011}=0\)
\(x+2018\cdot\left(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\right)=0\)
mà \(\left(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\right)\ne0\)
\(\Rightarrow x+2018=0\)
\(\Rightarrow x=-2018\)
Vậy,.............
Ta có: \(\frac{x+10}{2008}+\frac{x+9}{2009}=\frac{x+8}{2010}+\frac{x+7}{2011}\)
\(\Rightarrow\frac{x+10}{2008}+1+\frac{x+9}{2009}+1=\frac{x+8}{2010}+1+\frac{x+7}{2011}+1\)
\(\Rightarrow\frac{x+2018}{2008}+\frac{x+2018}{2009}=\frac{x+2018}{2010}+\frac{x+2018}{2011}\)
\(\Rightarrow\frac{x+2018}{2008}+\frac{x+2018}{2009}-\frac{x+2018}{2010}-\frac{x+2018}{2011}=0\)
\(\Rightarrow x+2018\cdot\left(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\right)=0\)
Do \(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\ne0\)
\(\Rightarrow x+2018=0\)
\(\Rightarrow x=-2018\)
Vậy \(x=-2018\)
Ta có:
\(\frac{x+7}{2010}+\frac{x+6}{2011}=\frac{x+5}{2012}+\frac{x+4}{2013}\)
\(\Leftrightarrow\left(\frac{x+7}{2010}+1\right)+\left(\frac{x+6}{2011}+1\right)=\left(\frac{x+5}{2012}+1\right)+\left(\frac{x+4}{2013}+1\right)\)
\(\Leftrightarrow\frac{x+2017}{2010}+\frac{x+2017}{2011}=\frac{x+2017}{2012}+\frac{x+2017}{2013}\)
\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2010}+\frac{1}{2011}\right)=\left(x+2017\right)\left(\frac{1}{2013}+\frac{1}{2014}\right)\)
Suy ra \(x+2017=0\)
Vậy \(x=-2017\)
b) Dễ tự làm nhé
x = 8 đúng luôn
làm thế nào vậy