Why x in equilibrium calculations can be "ignored"
In problems like this, the maths can sometimes be simplified by making assumptions so that you do not need to do difficult algebra. You could always do the harder algebra. 

In this case, a valid assumption is that when the equilibrium constant is very small in comparison to the concentrations given, then the change in those concentrations (which is represent by $x$) will also be very small and can be neglected. I'll work through your example the long way to show you why this assumption is valid. 

$$K_c= \dfrac{[\ce{ClF}][\ce{F2}]}{[\ce{ClF3}]} = \frac{(x)(1.00+x)}{(2.50-x)} = 8.77\times10^{-14}$$

Working through the algebra exactly:

$$(x)(1.00+x)=(8.77\times10^{-14})(2.50-x)$$
$$1.00x-x^2=2.19_{25}\times 10^{-13}-8.77\times10^{-14}x$$
$$0=x^2+(-8.77\times10^{-14}-1)x+2.19_{25}\times 10^{-13}$$

Solving the quadratic formula gives

$$x=\dfrac{(1+8.77\times10^{-14}\pm \sqrt{(-1-8.77\times10^{-14})^2 +4(2.19_{25}\times 10^{-13})}}{2}=[1,\ 2.19\times10^{-13}]$$

The first root $x = 1$ is nonsense. The second is the root you were given (when rounded to the correct number of sig figs).

The value of $x$ is so small compared to the initial concentrations of $\ce{ClF3}$ and $\ce{F2}$ that the concentrations of these species do not change within the magnitude of correct use of significant figures:

$$[\ce{ClF3}]_{eq}=2.50-2.1925\times10^{-13}=2.50 \ \text{to 2 decimal places}$$
$$[\ce{F2}]_{eq}=1.00+2.1925\times10^{-13}=1.00 \ \text{to 2 decimal places}$$

The maths above are equivalent to the assumption that since $x&lt;&lt;1.00$ and $x&lt;&lt;1.00$, $1.00+x\approx1.00$ and $2.50-x\approx2.50$. Whether you do it at the beginning or at the end, you get the same result: $x=2.19\times10^{-13}$. Making the simplification at the beginning is operationally easier and avoid potential algebra and arithmetic errors.

The assumption is not valid if the initial concentrations are close in value to $x$.