What is (2/(k+2)-1)/(2/(k+2)+1) ?

The solution to (2/(k+2)-1)/(2/(k+2)+1) is -k/(k+4)

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simplify (2/(k+2)-1)/(2/(k+2)+1)

simplify

\frac{\frac{2}{k + 2} - 1}{\frac{2}{k + 2} + 1}

- \frac{k}{k + 4}

Solution steps

\frac{\frac{2}{k + 2} - 1}{\frac{2}{k + 2} + 1}

Join

\frac{2}{k + 2} - 1 : - \frac{k}{k + 2}

= \frac{- \frac{k}{k + 2}}{\frac{2}{k + 2} + 1}

Join

\frac{2}{k + 2} + 1 : \frac{k + 4}{k + 2}

= \frac{- \frac{k}{k + 2}}{\frac{k + 4}{k + 2}}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{\frac{k}{k + 2}}{\frac{k + 4}{k + 2}}

Apply the fraction rule:

\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \cdot d}{b \cdot c}

\frac{\frac{k}{k + 2}}{\frac{k + 4}{k + 2}} = \frac{k ( k + 2 )}{( k + 2 ) ( k + 4 )}

= - \frac{k ( k + 2 )}{( k + 2 ) ( k + 4 )}

Cancel the common factor:

k + 2

= - \frac{k}{k + 4}

i^{23}

x^{2} + x = 90

3 (u + 3) + u = 2 (u - 1) + 3

(\frac{1}{x} + \frac{1}{x} + \frac{1}{x})^{- 1}

9 x^{2} - 3 x y + 6 x - 2 y

What is (2/(k+2)-1)/(2/(k+2)+1) ?
The solution to (2/(k+2)-1)/(2/(k+2)+1) is -k/(k+4)