The problem provides two real numbers $a$ and $b$ such that $a = 2 + \sqrt{3}$ and $ab = -\sqrt{3}$. We need to find $b$, calculate $3a^2 - b^2$, find the largest integer not exceeding $b$, and solve the inequality $bx > 3a - \frac{b^2}{a}$.

1. Problem Description

The problem provides two real numbers $a$ and $b$ such that $a = 2 + \sqrt{3}$ and $ab = -\sqrt{3}$.

We need to find $b$, calculate $3a^2 - b^2$, find the largest integer not exceeding $b$, and solve the inequality $bx > 3a - \frac{b^2}{a}$.

2. Solution Steps

(1)

From $ab = -\sqrt{3}$, we can find $b$ as

$b = \frac{-\sqrt{3}}{a} = \frac{-\sqrt{3}}{2 + \sqrt{3}}$.

To rationalize the denominator, we multiply the numerator and denominator by $2 - \sqrt{3}$:

$b = \frac{-\sqrt{3}(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{-2\sqrt{3} + 3}{4 - 3} = 3 - 2\sqrt{3}$.

So, $A = 3$, $B = 2$, and $C = 3$.

Now, we need to calculate $3a^2 - b^2$. We have $a = 2 + \sqrt{3}$ and $b = 3 - 2\sqrt{3}$.

$a^2 = (2 + \sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}$.

$3a^2 = 3(7 + 4\sqrt{3}) = 21 + 12\sqrt{3}$.

$b^2 = (3 - 2\sqrt{3})^2 = 9 - 12\sqrt{3} + 4(3) = 9 - 12\sqrt{3} + 12 = 21 - 12\sqrt{3}$.

$3a^2 - b^2 = (21 + 12\sqrt{3}) - (21 - 12\sqrt{3}) = 21 + 12\sqrt{3} - 21 + 12\sqrt{3} = 24\sqrt{3}$.

So, $DE = 0$ and $F = 24\sqrt{3}$.

Since $DE=0$, $3a^2 - b^2 = 24 \sqrt{3}$.

(2)

We need to find the largest integer not exceeding $b$.

$b = 3 - 2\sqrt{3} \approx 3 - 2(1.732) = 3 - 3.464 = -0.464$.

The largest integer not exceeding $b$ is $-1$.

So, $GH = -1$.

(3)

We need to solve the inequality $bx > 3a - \frac{b^2}{a}$.

$bx > 3a - \frac{b^2}{a} = \frac{3a^2 - b^2}{a} = \frac{24\sqrt{3}}{2 + \sqrt{3}}$.

Since $b = 3 - 2\sqrt{3} < 0$, we need to divide by $b$ and reverse the inequality sign:

$x < \frac{24\sqrt{3}}{a \cdot b} = \frac{24\sqrt{3}}{(2 + \sqrt{3})(3 - 2\sqrt{3})} = \frac{24\sqrt{3}}{6 - 4\sqrt{3} + 3\sqrt{3} - 6} = \frac{24\sqrt{3}}{-\sqrt{3}} = -24$.

Therefore, $x < -24$.

3. Final Answer

(1) $b = 3 - 2\sqrt{3}$. $3a^2 - b^2 = 24\sqrt{3}$.

(2) The largest integer not exceeding $b$ is $-1$.

(3) $x < -24$. The answer is ⑦.