1. The problem asks to find $x$ given the equation $\sqrt{5} + \sqrt{4} = \frac{1}{x}$. 2. First, simplify the square roots: $\sqrt{5}$ remains as is since 5 is not a perfect square, and $\sqrt{4} = 2$. 3. So the equation becomes $\sqrt{5} + 2 = \frac{1}{x}$. 4. To find $x$, take the reciprocal of both sides: $x = \frac{1}{\sqrt{5} + 2}$. 5. To rationalize the denominator, multiply numerator and denominator by the conjugate $2 - \sqrt{5}$: $$x = \frac{1}{\sqrt{5} + 2} \times \frac{2 - \sqrt{5}}{2 - \sqrt{5}} = \frac{2 - \sqrt{5}}{(\sqrt{5} + 2)(2 - \sqrt{5})}$$ 6. Multiply the denominator using the difference of squares formula: $$(\sqrt{5} + 2)(2 - \sqrt{5}) = 2^2 - (\sqrt{5})^2 = 4 - 5 = -1$$ 7. So, $$x = \frac{2 - \sqrt{5}}{-1} = -2 + \sqrt{5}$$ 8. Therefore, the solution is: $$x = -2 + \sqrt{5}$$