$\text { The function defined by } f(x)=\left\{\begin{array}{cc} \dfrac{\sin x}{x}+\cos x & x>0 \\ -5 k & x=0 \\ \dfrac{4(1-\sqrt{1-x})}{x} & x<0 \end{array} \quad \text { is continous at } x=0, \quad \text { then } k\right. \text { equals }$

For the function $f(x)$ to be continuous at $x = 0$, the left-hand limit, right-hand limit, and the value of the function at $x = 0$ must be equal.

The right-hand limit is given by $\lim_{x \to 0^{+}} f(x) = \lim_{x \to 0^{+}} \left( \dfrac{\sin x}{x} + \cos x \right) = 1 + 1 = 2$.

The left-hand limit is given by $\lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{-}} \dfrac{4(1 - \sqrt{1 - x})}{x}$.

Multiplying the numerator and denominator by the conjugate $(1 + \sqrt{1 - x})$:

$\lim_{x \to 0^{-}} \dfrac{4(1 - \sqrt{1 - x})(1 + \sqrt{1 - x})}{x(1 + \sqrt{1 - x})} = \lim_{x \to 0^{-}} \dfrac{4(1 - (1 - x))}{x(1 + \sqrt{1 - x})} = \lim_{x \to 0^{-}} \dfrac{4x}{x(1 + \sqrt{1 - x})} = \lim_{x \to 0^{-}} \dfrac{4}{1 + \sqrt{1 - x}} = \dfrac{4}{1 + 1} = 2$.

Since the limit exists and is equal to $2$, we set the value of the function at $x = 0$ equal to this limit:

$-5k = 2$.

Therefore, $k = -\dfrac{2}{5}$.

Answer: $-\dfrac{2}{5}$