Answer: [tex]\frac{14\text{x}+20}{105 \text{x}^2}\\\\[/tex]
Work Shown
[tex]\frac{2}{15 \text{x}}+\frac{4}{21 \text{x}^2}\\\\=\frac{7 \text{x}*2}{7 \text{x}*15 \text{x}}+\frac{5*4}{5*21 \text{x}^2}\\\\=\frac{14\text{x}}{105 \text{x}^2}+\frac{20}{105 \text{x}^2}\\\\=\frac{14\text{x}+20}{105 \text{x}^2}\\\\[/tex]
Therefore,
[tex]\frac{2}{15 \text{x}}+\frac{4}{21 \text{x}^2}=\frac{14\text{x}+20}{105 \text{x}^2}[/tex]
is an identity where [tex]\text{x} \ne 0[/tex]
In other words, it's a true equation for any nonzero value of x.
I used GeoGebra to confirm the answer.
