Determine the value range of rational number a, so that the system of inequalities about x 1 / 2x + X + 1 / 3 > 0 (1) x + 5A + 4 / 3 > 4 / 3 (x + 1) + a (2) has exactly two integer solutions

Determine the value range of rational number a, so that the system of inequalities about x 1 / 2x + X + 1 / 3 > 0 (1) x + 5A + 4 / 3 > 4 / 3 (x + 1) + a (2) has exactly two integer solutions

1\2x+ (x+1)\3＞0 x+ 5a+4\3＞4\3(x +1)+a
3x+2x+2＞0 3x+15a+4＞4x+4+3a
5x＞-2 -x＞-12a
x＞-2.5 x＜12a
∴-2.5＜12a
a＞-2.5/12
∴a1=1
a2=2

The inequality x ^ 2 + (a + 2) x + 3A + 6 > 0 holds for all real numbers, and the value range of a is obtained

Because x ^ 2 + (a + 2) x + 3A + 6 > 0 holds for all real numbers
So the discriminant of root is less than 0
And because the coefficient of x ^ 2 is greater than 0
So the image is a quadratic function above the x-axis
So (a + 2) ^ 2 - 4 (3a + 6) < 0
So a ^ 2 + 4A + 4-12a-24 < 0
a^2-8a-20＜0
（a-10）（a+2）＜0
So - 2 < a < 10

If x belongs to (- 3,1), the inequality x ^ 2-4ax + 3A ^ 2 > = 0 holds, the value range of real number a is obtained
Such as the title
Why does constant hold when a > 0

x^2-4ax+3a^2>=0
(x-a)(x-3a)>=0
If a > 0, when x belongs to (- 3,1), the inequality x ^ 2-4ax + 3A ^ 2 > = 0 holds (just draw the image of quadratic function)
If a = 0 x ^ 2 > = 0, it holds
If A0, when x is less than 0, the function is greater than 0

Try to determine the value range of real number a, so that the inequality system x2 + X + 13 ＞ 0x + 5A + 43 ＞ 43 (x + 1) + A has exactly two integer solutions

From x2 + X + 13 ＞ 0, multiply both sides by 6 to get 3x + 2 (x + 1) ＞ 0, the solution is x ＞ - 25, from x + 5A + 43 ＞ 43 (x + 1) + A, multiply both sides by 3 to get 3x + 5A + 4 ＞ 4 (x + 1) + 3a, the solution is x ＜ 2a, the solution set of the original inequality system is - 25 ＜ x ＜ 2A