Question

$\sqrt{2 {x}^{2} } + \sqrt[3]{ {2x}^{2} } + 6x \div 2x$

Answer 1

The question is not very clear. Perhaps 2x divides the entire expression, all terms.  So we need to simplify the terms.

$\frac{\sqrt{2x^2}+\sqrt[3]{2x^2}+6x}{2x}\\\\=\frac{(2x^2)^\frac{1}{2}+(2x^2)^\frac{1}{3}+6x}{2x}\\\\=2^{\frac{1}{2}-1} \: x^{2*\frac{1}{2}-1}+2^{\frac{1}{3}-1} * x^{2*\frac{1}{3}-1}+3 \\\\=\sqrt{2}*x^0+\frac{1}{\sqrt[3]{2^2 \: x}}+3$

Answer 2

Answer is in the image