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

\(\left| {x - 1} \right| = 1\)     

\(\left| x \right| =  - 9\)          

\(\left| {x + 3} \right| = 0\)

\(\left| {2x} \right| = 10\)

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

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

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

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

\(2x + 1 > 5\)

 \(7 - 2x < 10 - x\)

\(2 + x < 2 + 2x\)

\( - 3x > 4x + 3\)

\(x = \dfrac{3}{4};\,x = \dfrac{5}{4}\)          

\(x = \dfrac{1}{4};\,x = \dfrac{5}{4}\)             

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

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

\(x > 4 - 2\)

\(x >  - 4 + 2\)

\(x >  - 4 - 2\)

\(x > 4 + 2\)

\(\left\{ {\dfrac{5}{2}} \right\}\)       

\(\left\{ {\dfrac{5}{2};\dfrac{2}{3}} \right\}\)

\(\left\{ {\dfrac{5}{2}; - \dfrac{3}{2}} \right\}\)

\(\left\{ {\dfrac{5}{2}; - \dfrac{2}{3}} \right\}\)