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(A) Given: $f(x) = \frac{\sqrt{6x^2+5x-6}}{\sqrt{4-x}-\sqrt{x+4}}$For $f(x)$ to be defined:$(1)$ The numerator must be real: $6x^2+5x-6 \geq 0$$(3x-2)

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$[-4, -\frac{3}{2}] \cup [\frac{2}{3}, 4]$

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(A) Given: $f(x) = \frac{\sqrt{6x^2+5x-6}}{\sqrt{4-x}-\sqrt{x+4}}$For $f(x)$ to be defined:$(1)$ The numerator must be real: $6x^2+5x-6 \geq 0$$(3x-2)(2x+3) \geq 0 \Rightarrow x \in (-\infty, -\frac{3}{2}] \cup [\frac{2}{3}, \infty)$$(2)$ The denominator must be non-zero: $\sqrt{4-x} - \sqrt{x+4} 
eq 0$$4-x 
eq x+4$ $\Rightarrow 2x 
eq 0$ $\Rightarrow x 
eq 0$$(3)$ The square root terms must be defined:$4-x \geq 0 \Rightarrow x \leq 4$$x+4 \geq 0 \Rightarrow x \geq -4$Combining these conditions: $x \in [-4, 4] \cap ((-\infty, -\frac{3}{2}] \cup [\frac{2}{3}, \infty)) \cap \{x 
eq 0\}$Since $0$ is not in the interval $(-\infty, -\frac{3}{2}] \cup [\frac{2}{3}, \infty)$,the condition $x 
eq 0$ is automatically satisfied.Thus,the domain is $[-4, -\frac{3}{2}] \cup [\frac{2}{3}, 4]$.

The domain of the real valued function $f(x) = \frac{\sqrt{6x^2+5x-6}}{\sqrt{4-x}-\sqrt{x+4}}$ is

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