If A={x in R: sqrt{x^2-8x+15} in R} and B={x | vedclass

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(B) For set $A$, we require $x^2-8x+15 \geq 0$. $(x-3)(x-5) \geq 0$, which gives $x \in (-\infty, 3] \cup [5, \infty)$. For set $B$, we solve $\frac{x

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$\left(\frac{5}{2}, 3\right] \cup \left[5, \frac{11}{2}\right)$

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(B) For set $A$, we require $x^2-8x+15 \geq 0$. $(x-3)(x-5) \geq 0$, which gives $x \in (-\infty, 3] \cup [5, \infty)$. For set $B$, we solve $\frac{x-3}{2x-5} - \frac{x-6}{2x-11} $\frac{(x-3)(2x-11) - (x-6)(2x-5)}{(2x-5)(2x-11)} $\frac{(2x^2-17x+33) - (2x^2-17x+30)}{(2x-5)(2x-11)} $\frac{3}{(2x-5)(2x-11)} This implies $(2x-5)(2x-11) Finally, $A \cap B = ((-\infty, 3] \cup [5, \infty)) \cap \left(\frac{5}{2}, \frac{11}{2}\right) = \left(\frac{5}{2}, 3\right] \cup \left[5, \frac{11}{2}\right)$.

If $A=\{x \in R: \sqrt{x^2-8x+15} \in R\}$ and $B=\{x \in R: \frac{x-3}{2x-5} < \frac{x-6}{2x-11}\}$, then $A \cap B=$

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