Số nguyên nhỏ nhất thỏa mãn bất phương trình $x(5x + 1) + 4(x + 3) > 5{x^2}$ là

$5x + 7 < 0$                    

$0x + 6 > 0$

${x^2} - 2x > 0$                           

$x - 10 = 3$.

$7 - x < 2x$

$2x + 3 > 9$

$- 4x \ge x + 5$

$5 - x > 6x - 12$

$x > \dfrac{3}{5}$     

$x \le  - \dfrac{5}{3}$

$x \ge  - \dfrac{5}{3}$           

$x >  - \dfrac{5}{3}$.

$a - 3 > b - 3$

$- 3a + 4 >  - 3b + 4$

 $2a + 3 < 2b + 3$     

$- 5b - 1 <  - 5a - 1$.

$x = 3;x = 2$           

$x = \dfrac{5}{2};x = 2$      

$x = 1;x = 2$              

$x = 0,5{\rm{ }};x = 1,5$

$x = \dfrac{7}{{12}};x = \dfrac{1}{{12}}$          

$x = \dfrac{8}{{12}};x = \dfrac{5}{{12}}$             

$x = \dfrac{7}{{12}};x = \dfrac{2}{3}$          

$x = \dfrac{1}{{12}};x = \dfrac{5}{{12}}$

$x - 1 \ge 5$

$x + 1 \le 7$

$x + 3 < 9$

$x + 1 > 7$.