What is expand (k^a)/(n(k^a-n^a)) ?

The solution to expand (k^a)/(n(k^a-n^a)) is (k^a)/(nk^a-n^{1+a)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand (k^a)/(n(k^a-n^a))

expand

\frac{k ^{a}}{n ( k ^{a} - n ^{a} )}

\frac{k ^{a}}{n k ^{a} - n ^{1 + a}}

Solution steps

\frac{k ^{a}}{n ( k ^{a} - n ^{a} )}

Expand

n (k^{a} - n^{a}) : n k^{a} - n^{1 + a}

= \frac{k ^{a}}{n k ^{a} - n ^{a + 1}}

\frac{1}{( w i + 0.5 ) ^{2}}

(\frac{x}{2}) + 3

e^{x + 2} (2 x + 2)

\frac{( x + 5 )}{( x + 2 ) ( x + 3 ) ( x + 4 )}

β e (4 - θ^{2})^{2} - 2 t^{2} e

What is expand (k^a)/(n(k^a-n^a)) ?
The solution to expand (k^a)/(n(k^a-n^a)) is (k^a)/(nk^a-n^{1+a)}