Answer: (∞, 0) ∪ (0, 2/5)
In other words, x is less than 2/5 but cannot include 0
Step-by-step explanation:
[tex]f(x)\cdot g(x)=\dfrac{1}{x}\cdot\dfrac{1}{\sqrt{2-5x}}\\\\\\\text{The denominator cannot equal zero and must be a real number, so}\\x\neq 0\quad and\quad \sqrt{2-5x}>0\\\\\\(\sqrt{2-5x})^2>0^2\\.\quad 2-5x>0\\.\qquad -5x>-2\\\\.\qquad \quad x<\dfrac{-2}{-5}\\\\.\qquad \quad x<\dfrac{2}{5}\\\\\\\\\text{So, }x<\dfrac{2}{5}\quad \text{but cannot equal 0}\\\\\\\text{Interval Notation: }\large\boxed{ \bigg(-\infty,0\bigg) \text{U}\bigg(0, \dfrac{2}{5}\bigg)}[/tex]
