What is (2j^{-1}+5j^{-2})/(4-25j^{-2)} ?

The solution to (2j^{-1}+5j^{-2})/(4-25j^{-2)} is 1/(2j-5)

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simplify (2j^{-1}+5j^{-2})/(4-25j^{-2)}

simplify

\frac{2 j ^{- 1} + 5 j ^{- 2}}{4 - 25 j ^{- 2}}

\frac{1}{2 j - 5}

Solution steps

\frac{2 j ^{- 1} + 5 j ^{- 2}}{4 - 25 j ^{- 2}}

25 j^{- 2} = \frac{25}{j ^{2}}

= \frac{2 j ^{- 1} + 5 j ^{- 2}}{4 - \frac{25}{j ^{2}}}

2 j^{- 1} = \frac{2}{j}

5 j^{- 2} = \frac{5}{j ^{2}}

= \frac{\frac{2}{j} + \frac{5}{j ^{2}}}{4 - \frac{25}{j ^{2}}}

Join

4 - \frac{25}{j ^{2}} : \frac{4 j ^{2} - 25}{j ^{2}}

= \frac{\frac{2}{j} + \frac{5}{j ^{2}}}{\frac{4 j ^{2} - 25}{j ^{2}}}

Join

\frac{2}{j} + \frac{5}{j ^{2}} : \frac{2 j + 5}{j ^{2}}

= \frac{\frac{2 j + 5}{j ^{2}}}{\frac{4 j ^{2} - 25}{j ^{2}}}

Divide fractions:

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

= \frac{( 2 j + 5 ) j ^{2}}{j ^{2} ( 4 j ^{2} - 25 )}

Cancel the common factor:

j^{2}

= \frac{2 j + 5}{4 j ^{2} - 25}

Factor

4 j^{2} - 25 : (2 j + 5) (2 j - 5)

= \frac{2 j + 5}{( 2 j + 5 ) ( 2 j - 5 )}

Cancel the common factor:

2 j + 5

= \frac{1}{2 j - 5}

2 x^{3} + x^{2} - 13 x + 6

a^{2} + 2 a - 8

(p + q)^{2} + 4 (p + q) + 4

x^{2} - \frac{49}{25}

2 x (4 - x) - \frac{3}{5} (15 x - 7) = \frac{1}{2} (4 - 6 x) - 2 x^{2}

What is (2j^{-1}+5j^{-2})/(4-25j^{-2)} ?
The solution to (2j^{-1}+5j^{-2})/(4-25j^{-2)} is 1/(2j-5)