What is (\frac{v^2-9)/(3v^2+12)}{(v+3)/(v^2+4)} ?

The solution to (\frac{v^2-9)/(3v^2+12)}{(v+3)/(v^2+4)} is (v-3)/3

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simplify (\frac{v^2-9)/(3v^2+12)}{(v+3)/(v^2+4)}

simplify

\frac{\frac{v ^{2} - 9}{3 v ^{2} + 12}}{\frac{v + 3}{v ^{2} + 4}}

\frac{v - 3}{3}

Solution steps

\frac{\frac{v ^{2} - 9}{3 v ^{2} + 12}}{\frac{v + 3}{v ^{2} + 4}}

Apply the fraction rule:

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

= \frac{( v ^{2} - 9 ) ( v ^{2} + 4 )}{( 3 v ^{2} + 12 ) ( v + 3 )}

Cancel

\frac{( v ^{2} - 9 ) ( v ^{2} + 4 )}{( 3 v ^{2} + 12 ) ( v + 3 )} : \frac{v - 3}{3}

= \frac{v - 3}{3}

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

\frac{x ^{2} - 16}{x ^{2} + 9 x + 20}

lo g_{10} (1 0^{- 6})

\frac{10}{10}

- 64 - 12 x^{2} + 48 x + x^{3}

What is (\frac{v^2-9)/(3v^2+12)}{(v+3)/(v^2+4)} ?
The solution to (\frac{v^2-9)/(3v^2+12)}{(v+3)/(v^2+4)} is (v-3)/3