Solve the inequality:
1. x^2 - (2/x) > 0
2. (x-5)^4 / (x(x+3)) ≥ 0
Thank you for any help!
$\displaystyle x^{2}-\frac{2}{x}=\frac{x^{3}-2}{x}=\frac{\left( x-\sqrt[3]{2} \right)\left( x^{2}+\sqrt[3]{2}x+\sqrt[3]{4} \right)}{x}>0,$ and since $\displaystyle x^{2}+\sqrt[3]{2}x+\sqrt[3]{4}>0,$ you just need to solve $\displaystyle \frac{x-\sqrt[3]{2}}{x}>0.$
Numerator is always positive (and zero when $\displaystyle x=5$), so, you just need to solve $\displaystyle x(x+3)>0.$