Tìm $x$ để biểu thức sau có giá trị dương $A = \dfrac{{x + 27}}{5} - \dfrac{{3x - 7}}{4}$

$\left| {x - 1} \right| = 1$     

$\left| x \right| =  - 9$          

$\left| {x + 3} \right| = 0$

$\left| {2x} \right| = 10$

$S = \left\{ x \in R|{x \ge \dfrac{1}{2}} \right\}$

$S = \left\{ x \in R|{x \ge  - \dfrac{1}{2}} \right\}$

$S = \left\{ x \in R|{x \le  - \dfrac{1}{2}} \right\}$

$S = \left\{ x \in R|{x \le \dfrac{1}{2}} \right\}$

$2x + 1 > 5$

 $7 - 2x < 10 - x$

$2 + x < 2 + 2x$

$- 3x > 4x + 3$

$x = \dfrac{3}{4};\,x = \dfrac{5}{4}$          

$x = \dfrac{1}{4};\,x = \dfrac{5}{4}$             

$x =  - \dfrac{1}{4};\,x = \dfrac{5}{4}$          

$x = \dfrac{1}{4};\,x =  - \dfrac{5}{4}$

$x > 4 - 2$

$x >  - 4 + 2$

$x >  - 4 - 2$

$x > 4 + 2$

$\left\{ {\dfrac{5}{2}} \right\}$       

$\left\{ {\dfrac{5}{2};\dfrac{2}{3}} \right\}$

$\left\{ {\dfrac{5}{2}; - \dfrac{3}{2}} \right\}$

$\left\{ {\dfrac{5}{2}; - \dfrac{2}{3}} \right\}$