a, \(2x-3< 0\Leftrightarrow2x< 3\Leftrightarrow x< \frac{3}{2}\)

b, \(\left(2x-4\right)\left(9-3x\right)>0\)

\(\Leftrightarrow\hept{\begin{cases}2x-4>0\\9-3x>0\end{cases}\Leftrightarrow\hept{\begin{cases}x>2\\x< 3\end{cases}\Leftrightarrow2< x< 3}}\)

a. \(2x-3< 0\Leftrightarrow2x< 3\Leftrightarrow x< \frac{3}{2}\)

b. \(\left(2x-4\right)\left(9-3x\right)>0\Leftrightarrow18x-6x-36+12x>0\Leftrightarrow24x>36\Leftrightarrow x>\frac{3}{2}\)

c. \(\frac{2}{3}x-\frac{3}{4}>0\Leftrightarrow\frac{2}{3}x>\frac{3}{4}\Leftrightarrow x>\frac{9}{8}\)

d. \(\left(\frac{3}{4}-2x\right)\left(\frac{-3}{5}+\frac{2}{-61}-\frac{17}{51}\right)\le0\)

\(\Leftrightarrow\frac{3}{4}-2x\le0\Leftrightarrow2x\le\frac{3}{4}\Leftrightarrow x\le\frac{3}{8}\)

e. \(\left(\frac{3}{2}x-4\right).\frac{5}{3}>\frac{15}{6}\Leftrightarrow\frac{3}{2}x-4>\frac{3}{2}\Leftrightarrow\frac{3}{2}x>\frac{11}{2}\Leftrightarrow x>\frac{11}{3}\)

a: \(x+\dfrac{3}{9}=\dfrac{7}{6}\cdot\dfrac{2}{3}\)

=>\(x+\dfrac{1}{3}=\dfrac{14}{18}=\dfrac{7}{9}\)

=>\(x=\dfrac{7}{9}-\dfrac{1}{3}=\dfrac{7}{9}-\dfrac{3}{9}=\dfrac{4}{9}\)

b: \(x-\dfrac{2}{3}=\dfrac{1}{8}:\dfrac{5}{4}\)

=>\(x-\dfrac{2}{3}=\dfrac{1}{8}\cdot\dfrac{4}{5}=\dfrac{1}{10}\)

=>\(x=\dfrac{1}{10}+\dfrac{2}{3}=\dfrac{3+20}{30}=\dfrac{23}{30}\)

TThế giới oi oi oi 

\(a,x=\dfrac{13}{2}-2\\ x=\dfrac{9}{2}\\ b,x=\dfrac{4}{5}\times\dfrac{3}{4}\\ x=\dfrac{12}{20}=\dfrac{3}{5}\)

KHO THE

\(A=\frac{\left[\left(25-1\right):1+1\right]\left(25+1\right)}{2}=325.\)

\(B=\frac{\left[\left(51-3\right):2+1\right]\left(51+3\right)}{2}=675\)

\(C=\frac{\left[\left(81-1\right):4+1\right]\left(81+1\right)}{2}=861\)

Ko cần đâu bn à mk mong bn đấy

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

\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5-\frac{1}{2}x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)

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

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

    \(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{8}\)

\(\Rightarrow\hept{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{29}{12}\\x=-\frac{13}{12}\end{cases}}\)

a)\(\left(3x-1\right)\left(\frac{-1}{2}x+5\right)=0\)
\(\Leftrightarrow\)3x - 1 = 0      hay      \(\frac{-1}{2}\)x + 5 = 0
\(\Leftrightarrow\)3x     = 1         I\(\Leftrightarrow\)\(\frac{-1}{2}\)x     = -5
\(\Leftrightarrow\)  x     = \(\frac{1}{3}\)  I\(\Leftrightarrow\)            x     = 10

b) 2 I \(\frac{1}{2}x-\frac{1}{3}\)I - \(\frac{3}{2}\)=\(\frac{1}{4}\)
\(\Leftrightarrow\) 2 I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{4}\)
\(\Leftrightarrow\)    I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x-\frac{1}{3}\)= \(\frac{7}{8}\)          hay     \(\frac{1}{2}x-\frac{1}{3}\)= \(\frac{-7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x\)           = \(\frac{29}{24}\)        I\(\Leftrightarrow\)\(\frac{1}{2}x\)           = \(\frac{-13}{24}\)
\(\Leftrightarrow\)      x              = \(\frac{29}{12}\)        I\(\Leftrightarrow\)      x              = \(\frac{-13}{12}\)

c) (2x +\(\frac{3}{5}\))² - \(\frac{9}{25}\)= 0
\(\Leftrightarrow\)(2x +\(\frac{3}{5}\))²       = \(\frac{9}{25}\)
\(\Leftrightarrow\) 2x +\(\frac{3}{5}\)         = \(\frac{3}{5}\)    hay      2x +\(\frac{3}{5}\)= \(\frac{-3}{5}\)
\(\Leftrightarrow\) 2x                    = 0           I \(\Leftrightarrow\)2x           = \(\frac{-6}{5}\)
\(\Leftrightarrow\)   x                    = 0           I \(\Leftrightarrow\) x           = \(\frac{-3}{5}\)

d) 3(x -\(\frac{1}{2}\)) - 5(x +\(\frac{3}{5}\)) = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)3x - \(\frac{3}{2}\)- 5x - 3 = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)-2x + x - \(\frac{9}{2}\)- \(\frac{1}{5}\)= 0
\(\Leftrightarrow\)-x = \(\frac{-47}{10}\)
\(\Leftrightarrow\) x = \(\frac{47}{10}\)

a)\(3x-\dfrac{2}{5}=0=>3x=\dfrac{2}{5}=>x=\dfrac{2}{15}\)

b)\(\left(x-3\right)\left(2x+8\right)=0=>\left[{}\begin{matrix}x-3=0\\2x=-8\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)

c)\(3x^2-x-4=0=>3x^2+3x-4x-4=0=>\left(3x-4\right)\left(x+1\right)=0\)

\(=>\left[{}\begin{matrix}3x=4\\x+1=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-1\end{matrix}\right.\)

\(\dfrac{1}{2}-\dfrac{5}{12}x=\dfrac{2}{3}\)

\(\dfrac{5}{12}x=\dfrac{1}{2}-\dfrac{2}{3}=\dfrac{3}{6}-\dfrac{4}{6}\)

\(\dfrac{5}{12}x=\dfrac{-1}{6}\)

\(x=\dfrac{-1}{6}:\dfrac{5}{12}=\dfrac{-1}{6}.\dfrac{12}{5}\)

\(x=\dfrac{-2}{5}\)

Giúp mik ik 

\(\dfrac{1}{2}-\dfrac{5}{12}x=\dfrac{2}{3}\)

\(\dfrac{5}{12}x=\dfrac{-1}{6}\)

\(x=\dfrac{-2}{5}\)