To put
[tex]\frac{5}{7-i}[/tex]In the standard form, we must multiply the numerator and denominator by the complex conjugate of (7-i), it means
[tex]\frac{5}{7-i}\cdot\frac{7+i}{7+i}[/tex]And now we solve it, therefore
[tex]\frac{5}{7-i}\cdot\frac{7+i}{7+i}=\frac{5(7+i)}{7^2-i^2}[/tex]Remember that
[tex]i^2=-1[/tex]Then
[tex]\begin{gathered} \frac{5(7+i)}{7^2-i^2}=\frac{5(7+i)}{49+1} \\ \\ \frac{5(7+i)}{49+1}=\frac{5(7+i)}{50} \end{gathered}[/tex]Now we can simplify it
[tex]\frac{5(7+i)}{50}=\frac{7+i}{10}[/tex]And we have it in the standard form
[tex]\frac{7}{10}+\frac{1}{10}i[/tex]