What is (\frac{2a+1)/(10a-5)}{(10a)/(4a^2-1)} ?

The solution to (\frac{2a+1)/(10a-5)}{(10a)/(4a^2-1)} is ((2a+1)^2)/(50a)

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simplify (\frac{2a+1)/(10a-5)}{(10a)/(4a^2-1)}

simplify

\frac{\frac{2 a + 1}{10 a - 5}}{\frac{10 a}{4 a ^{2} - 1}}

\frac{( 2 a + 1 ) ^{2}}{50 a}

Solution steps

\frac{\frac{2 a + 1}{10 a - 5}}{\frac{10 a}{4 a ^{2} - 1}}

Apply the fraction rule:

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

= \frac{( 2 a + 1 ) ( 4 a ^{2} - 1 )}{( 10 a - 5 ) \cdot 10 a}

Cancel

\frac{( 2 a + 1 ) ( 4 a ^{2} - 1 )}{( 10 a - 5 ) \cdot 10 a} : \frac{( 2 a + 1 ) ^{2}}{50 a}

= \frac{( 2 a + 1 ) ^{2}}{50 a}

- x (5 x - 4)

15 \cdot 20

\frac{1}{1 + x ^{2}} d x

\frac{4 x ^{2} - 9}{2 x ^{2} - x - 3}

What is (\frac{2a+1)/(10a-5)}{(10a)/(4a^2-1)} ?
The solution to (\frac{2a+1)/(10a-5)}{(10a)/(4a^2-1)} is ((2a+1)^2)/(50a)