( x - 3/2 ) ( 2x + 1 ) > 0

TH1 : cả 2 thừa số đều lớn hơn 0

\(\Rightarrow\hept{\begin{cases}x-\frac{3}{2}>0\\2x+1>0\end{cases}\Rightarrow\hept{\begin{cases}x>\frac{3}{2}\\x>-\frac{1}{2}\end{cases}\Rightarrow}x>\frac{3}{2}}\)

TH2 : cả 2 thừa số đều bé hơn 0

\(\Rightarrow\hept{\begin{cases}x-\frac{3}{2}< 0\\2x+1< 0\end{cases}\Rightarrow\hept{\begin{cases}x< \frac{3}{2}\\x< -\frac{1}{2}\end{cases}\Rightarrow}x< -\frac{1}{2}}\)

Vậy,..........

\(a,\dfrac{-5}{x-3}< 0\Leftrightarrow x-3>0\left(-5< 0\right)\Leftrightarrow x>3\\ b,\dfrac{3-x}{x^2+1}\ge0\Leftrightarrow3-x\ge0\left(x^2+1>0\right)\Leftrightarrow x\le3\\ c,\dfrac{\left(x-1\right)^2}{x-2}< 0\Leftrightarrow x-2< 0\left[\left(x-1\right)^2\ge0\right]\Leftrightarrow x< 2\)

a) \(\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)

=> \(\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x+\frac{1}{3}=0\)

=> \(\left(\frac{2}{3}x-\frac{1}{2}x\right)+\left(-\frac{2}{5}+\frac{1}{3}\right)=0\)

=> \(\frac{1}{6}x-\frac{1}{15}=0\Rightarrow\frac{1}{6}x=\frac{1}{15}\Rightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{2}{5}\)

Vậy x = 2/5

b) \(\frac{1}{3}x+\frac{2}{5}\left(x+1\right)=0\)

=> \(\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)

=> \(\frac{11}{15}x+\frac{2}{5}=0\Rightarrow\frac{11}{15}x=-\frac{2}{5}\)

=> \(x=\left(-\frac{2}{5}\right):\frac{11}{15}=\left(-\frac{2}{5}\right)\cdot\frac{15}{11}=-\frac{6}{11}\)

Vậy x = -6/11

c) \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)

=> \(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)

=> \(\left(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}\right)+\left(-\frac{1}{3}x-x\right)=5\)

=> \(\frac{2}{3}-\frac{4}{3}x=5\)

=> \(\frac{4}{3}x=-\frac{13}{3}\Rightarrow x=\left(-\frac{13}{3}\right):\frac{4}{3}=\left(-\frac{13}{3}\right)\cdot\frac{3}{4}=-\frac{13}{4}\)

Vậy x = -13/4

d) \(\frac{11}{5}-\left(\frac{7}{9}-x\right)\cdot\frac{3}{8}=\frac{61}{90}+\frac{x}{3}\)

=> \(\frac{11}{5}-\frac{3}{8}\left(\frac{7}{9}-x\right)=\frac{61}{90}+\frac{30x}{90}\)

=> \(\frac{11}{5}-\frac{7}{24}+\frac{3}{8}x=\frac{61+30x}{90}\)

=> \(\frac{229}{120}+\frac{3}{8}x=\frac{61+30x}{90}\)

=> \(\frac{229}{120}+\frac{3x}{8}=\frac{61+30x}{90}\)

=> \(\frac{229}{120}+\frac{45x}{120}=\frac{61+30x}{90}\)

=> \(\frac{229+45x}{120}=\frac{61+30x}{90}\)

=> \(\frac{3\left(229+45x\right)}{360}=\frac{4\left(61+30x\right)}{360}\)

=> \(3\left(229+45x\right)=4\left(61+30x\right)\)

=> \(687+135x=244+120x\)

=> \(687+135x-244-120x=0\)

=> \(\left(687-244\right)+\left(135x-120x\right)=0\)

=> \(443+15x=0\)

=> \(15x=-443\Rightarrow x=-\frac{443}{15}\)

Vậy x = -443/15

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

Câu 1 :

\(a,2\left(\frac{3}{4}-5x\right)=\frac{4}{5}-3x\)

\(\Rightarrow\frac{3}{2}-10x=\frac{4}{5}-3x\)

\(\Rightarrow7x=\frac{3}{2}-\frac{4}{5}\)

\(\Rightarrow7x=\frac{7}{10}\)\(\Leftrightarrow x=0,1\)

\(b,\frac{3}{2}-4\left(\frac{1}{4}-x\right)=\frac{2}{3}-7x\)

\(\Rightarrow\frac{3}{2}-1+4x=\frac{2}{3}-7x\)

\(\Rightarrow11x=\frac{2}{3}+1-\frac{3}{2}\)

\(\Rightarrow11x=\frac{4+6-9}{6}-\frac{1}{6}\)

\(\Rightarrow x=\frac{1}{66}\)

Câu 2 :

\(a,\frac{2}{x-1}< 0\)

Vì \(2>0\Rightarrow\)để \(\frac{2}{x-1}< 0\)thì \(x-1< 0\Leftrightarrow x< 1\)

\(b,\frac{-5}{x-1}< 0\)

Vì \(-5< 0\)\(\Rightarrow\)để \(\frac{-5}{x-1}< 0\)thì \(x-1>0\Rightarrow x>1\)

\(c,\frac{7}{x-6}>0\)

Vì \(7>0\Rightarrow\)để \(\frac{7}{x-6}>0\)thì \(x-6>0\Rightarrow x>6\)

B1:

a)\(2\left(\frac{3}{4}-5x\right)=\frac{4}{5}-3x\) 

 \(\frac{3}{2}-10x=\frac{4}{5}-3x\) 

    \(3x-10x=\frac{4}{5}-\frac{3}{2}\) 

    \(-7x=\frac{-7}{10}\)

          \(x=10\) 

Vậy.........

b)\(\frac{3}{2}-4\left(\frac{1}{4}-x\right)=\frac{2}{3}-7x\) 

    \(\frac{3}{2}-4+4x=\frac{2}{3}-7x\) 

      \(\frac{-5}{2}+4x=\frac{2}{3}-7x\) 

          \(7x+4x=\frac{2}{3}+\frac{5}{2}\) 

             \(11x=\frac{19}{6}\) 

                \(x=\frac{19}{66}\) 

Vậy.........