What is 1/(\frac{3){x^2-4}-3} ?

The solution to 1/(\frac{3){x^2-4}-3} is (x^2-4)/(-3x^2+15)

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simplify 1/(\frac{3){x^2-4}-3}

simplify

\frac{1}{\frac{3}{x ^{2} - 4} - 3}

\frac{x ^{2} - 4}{- 3 x ^{2} + 15}

Solution steps

\frac{1}{\frac{3}{x ^{2} - 4} - 3}

Join

\frac{3}{x ^{2} - 4} - 3 : \frac{- 3 x ^{2} + 15}{x ^{2} - 4}

= \frac{1}{\frac{- 3 x ^{2} + 15}{x ^{2} - 4}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{x ^{2} - 4}{- 3 x ^{2} + 15}

tan^{2} (x) - tan (x)

\frac{12}{5} + 1

2 x - x^{2} - 1

224 - 554 + 12600

+ 64 x^{3} - 125

What is 1/(\frac{3){x^2-4}-3} ?
The solution to 1/(\frac{3){x^2-4}-3} is (x^2-4)/(-3x^2+15)