Bài 1:

Tìm x, biết:

$\begin{array}{l}a)2x + 3 = 1\dfrac{2}{3}\\b)0,15 - 3x = {( - 10)^0}\\c) - x:\dfrac{2}{5} = 0,8\\d)\dfrac{{3x + 2}}{3} = \dfrac{{ - 4}}{5}\\e)\dfrac{{3x + 2}}{{ - 8}} = \dfrac{{ - 2}}{{3x + 2}}\\f)\left( {x + 1} \right).\left( { - 2x - 3} \right) = 0\end{array}$

Bài 2:

Tìm $x$, biết:

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

b) $3 \cdot {\left( {3x - \dfrac{1}{2}} \right)^3} + \dfrac{1}{9} = 0$

c) $3 \cdot \left( {1 - \dfrac{1}{2}} \right) - 5 \cdot \left( {x + \dfrac{3}{5}} \right) = {\rm{ \;}} - x + \dfrac{1}{5}$                                     

d) $\dfrac{{3 - x}}{{5 - x}} = {\left( {\dfrac{{ - 3}}{5}} \right)^2}$

e) $x\;:\;\dfrac{5}{8} = \dfrac{{ - 13}}{{35}} \cdot \dfrac{{15}}{{ - 39}}$                                                           

f) $\left( {\dfrac{7}{5}\; + \;x} \right):\dfrac{{25}}{{16}} = \dfrac{{ - 4}}{5}$

g) $- 4:\left( {x + \dfrac{{ - 2}}{3}} \right) = \dfrac{3}{4}$                                                                

h) $\left( {\dfrac{{ - 1}}{5} + 2} \right):\left( {x - \dfrac{7}{{10}}} \right) = \dfrac{{ - 1}}{4}$

Bài 3:

Tìm tập hợp các số nguyên x để: $\dfrac{5}{6} + \dfrac{{ - 7}}{8} \le \dfrac{x}{{24}} \le \dfrac{{ - 5}}{{12}} + \dfrac{5}{8}$