Answer :

[tex]\frac{16+x}{x^3}+\frac{7-4x}{x^3}=\frac{16+x+7-4x}{x^3}=\frac{23-3x}{x^3}\\\\======================================\\\\\frac{5}{t-1}+\frac{3}{t}=\frac{5t}{t(t-1)}+\frac{3(t-1)}{t(t-1)}=\frac{5t+3t-3}{t^2-t}=\frac{8t-3}{t^2-t}[/tex]
MathG33k
[tex] \frac{16+x}{x^{3} } + \frac{7-4x}{ x^{3} } \\ 16+x+7-4x = 23-3x \\ \frac{23 -3x}{ x^{3} } [/tex]
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[tex] \frac{5}{t-1} +\frac{3}{t} \\ \\ \frac{5t+(t-1)(3)}{(t-1)(t)} \\ \\ \frac{5t+3t-3}{ t^{2}-t } \\ \\ \frac{8t-3}{ t^{2}-t } [/tex]