What is expand ((e^8-1)^2)/(4e^8)+1 ?

The solution to expand ((e^8-1)^2)/(4e^8)+1 is (e^8)/4-1/2+1/(4e^8)+1

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expand ((e^8-1)^2)/(4e^8)+1

expand

\frac{( e ^{8} - 1 ) ^{2}}{4 e ^{8}} + 1

\frac{e ^{8}}{4} - \frac{1}{2} + \frac{1}{4 e ^{8}} + 1

Solution steps

\frac{( e ^{8} - 1 ) ^{2}}{4 e ^{8}} + 1

(e^{8} - 1)^{2} = e^{16} - 2 e^{8} + 1

= \frac{e ^{16} - 2 e ^{8} + 1}{4 e ^{8}} + 1

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{e ^{16} - 2 e ^{8} + 1}{4 e ^{8}} = \frac{e ^{16}}{4 e ^{8}} - \frac{2 e ^{8}}{4 e ^{8}} + \frac{1}{4 e ^{8}}

= \frac{e ^{16}}{4 e ^{8}} - \frac{2 e ^{8}}{4 e ^{8}} + \frac{1}{4 e ^{8}} + 1

Cancel

\frac{e ^{16}}{4 e ^{8}} : \frac{e ^{8}}{4}

= \frac{e ^{8}}{4} - \frac{2 e ^{8}}{4 e ^{8}} + \frac{1}{4 e ^{8}} + 1

Cancel

\frac{2 e ^{8}}{4 e ^{8}} : \frac{1}{2}

= \frac{e ^{8}}{4} - \frac{1}{2} + \frac{1}{4 e ^{8}} + 1

e^{t} u (t - 1)

\frac{( y - ( ( x + 1 ) \cdot 4 ) )}{x}

x + 2 y - 4.2 x + y - 5

A (3, 5) y B (5, - 3)

(1, 0, 1) x (2, 1, - 1)

What is expand ((e^8-1)^2)/(4e^8)+1 ?
The solution to expand ((e^8-1)^2)/(4e^8)+1 is (e^8)/4-1/2+1/(4e^8)+1