What is expand b/(n(b-g-u))+c/(e^{t(b-g-u))} ?

The solution to expand b/(n(b-g-u))+c/(e^{t(b-g-u))} is b/(nb-ng-nu)+c/(e^{bt-gt-ut)}

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expand b/(n(b-g-u))+c/(e^{t(b-g-u))}

expand

\frac{b}{n ( b - g - u )} + \frac{c}{e ^{t (b - g - u)}}

\frac{b}{nb - n g - n u} + \frac{c}{e ^{b t - g t - u t}}

Solution steps

\frac{b}{n ( b - g - u )} + \frac{c}{e ^{t (b - g - u)}}

Expand

n (b - g - u) : nb - n g - n u

= \frac{b}{nb - n g - n u} + \frac{c}{e ^{t (b - g - u)}}

e^{t (b - g - u)} = e^{b t - g t - u t}

= \frac{b}{nb - n g - n u} + \frac{c}{e ^{b t - g t - u t}}

- \frac{4 ( 9 y ^{2} + 4 x ^{2} )}{81 y ^{3}}

\frac{1}{( x - a ) ^{3}} - \frac{1}{( x + a ) ^{2}}

\frac{s + 1}{( s ^{2} + 2 s + 10 ) ^{2}}

f^{'} (1) (x - 1)

\frac{( n + 8 ) ( n + 1 )}{n ( n - 10 )}

What is expand b/(n(b-g-u))+c/(e^{t(b-g-u))} ?
The solution to expand b/(n(b-g-u))+c/(e^{t(b-g-u))} is b/(nb-ng-nu)+c/(e^{bt-gt-ut)}