Answer:
[tex]\frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}[/tex]
Step-by-step explanation:
[tex]\frac{\frac{1}{c}+\frac{1}{h}}{\frac{1}{c^2}-\frac{1}{r^2}}[/tex]
Combine [tex]\frac{1}{c} + \frac{1}{h}[/tex]
[tex]\frac{\frac{h+c}{ch}}{\frac{1}{c^2}-\frac{1}{r^2}}[/tex]
Combine the bottom, too.
[tex]=\frac{\frac{h+c}{ch}}{\frac{r^2-c^2}{c^2r^2}}[/tex]
Apply the fraction rule
[tex]=\frac{\left(h+c\right)c^2r^2}{ch\left(r^2-c^2\right)}[/tex]
Cancel
[tex]=\frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}[/tex]
Therefore, [tex]\frac{\left(\frac{1}{c}+\frac{1}{h}\right)}{\left(\frac{1}{\left(c^2\right)}-\frac{1}{\left(r^2\right)}\right)}:\quad \frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}[/tex]
