Easy product :)

Factorise $x$ as:

$x = (1+\frac{1}{2})(1+\frac{1}{3})...(1 + \frac{1}{15})\cdot(1-\frac{1}{2})(1-\frac{1}{3})...(1 - \frac{1}{15})$

Simplifying each terms and we will get:

$\Large x = \frac{3 \cdot 4 \cdot 5 \cdot ... \cdot 16}{2 \cdot 3 \cdot 4 \cdot 5 \cdot...\cdot 15} \cdot \frac{1 \cdot 2 \cdot 3 \cdot...\cdot14}{2 \cdot 3 \cdot 4 \cdot...\cdot 15}$

Note that the numerator and the denominator will just cancel out.

$x = 8 \cdot \frac{1}{15} = \frac{8}{15}$

Thus $a + b = \boxed{23}$

Using the identity $A^{2} - B^{2} = (A-B)(A+B)$ will be a lot easier. You can cancel out the numbers to get the answer easily

$x= (1- \frac{1}{2^2})(1- \frac{1}{3^2})\ldots (1-\frac{1}{15^2})$
$\Rightarrow x= (\frac{2^2-1}{2^2}) (\frac{3^2-1}{3^2}) \ldots (\frac{15^2-1}{15^2})$
$\Rightarrow x= (\frac{(2+1)(2-1)}{2^2}) (\frac{(3+1)(3-1)}{3^2}) \ldots (\frac{(15+1)(15-1)}{15^2})$
$\Rightarrow x= (\frac{(3 \times 1)( 4 \times 2) \ldots (16 \times 14)}{2^2 3^2 \ldots 15^2}$
$\Rightarrow x= (\frac { 1\times 2 \times (3^2 4^2 \ldots 14^2) \times 15 \times 16}{2^2 \times (3^2 4^2 \ldots 14^2) \times 15^2}$
$\Rightarrow x= \frac{1 \times 2 \times 15 \times 16}{2^2 \times 15^2}$
$\Rightarrow x= \frac{8}{15}$
So, $a=8$ & $b=15$ .
Hence, $a+b=\boxed {23}$

As Cheah Chung Yin has said, using the identity $A^2-B^2=(A-B)(A+B)$ , we can turn the equation into $(1^2-(\frac{1}{2})^2)(1^2-(\frac{1}{3})^2)\ldots(1^2-(\frac{1}{15})^2)$ which equals $(1-\frac{1}{2})(1+\frac{1}{2})(1-\frac{1}{3})(1+\frac{1}{3})\ldots(1-\frac{1}{2})(1+\frac{1}{15})$ . Regroup to get $(\frac{1}{2})(\frac{2}{3})\ldots(\frac{14}{15})\times(\frac{3}{2})(\frac{4}{3})\ldots(\frac{16}{15})$ . Then, everything cancels out and you get $\frac{1}{15}\times\frac{16}{2}=\frac{8}{15}$ . Then, adding 8 and 15, we get the final answer of $\boxed{23}$ .