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Simplify. Is this correct?

My answer = c^2 + d^2

Guest Mar 10, 2015

Best Answer 

 #3
avatar+18845 
+10

$$\small{\text{
$
\dfrac{
\dfrac{c}{d}-\dfrac{d}{c}
}
{
\dfrac{1}{c}-\dfrac{1}{d}
}
=
\dfrac{
\dfrac{c^2-d^2}{dc}
}
{
\dfrac{d-c}{dc}
}
= \left( \dfrac{c^2-d^2}{dc} \right) \cdot
\left( \dfrac{dc}{d-c} \right)
= \dfrac{c^2-d^2}{d-c}
= -\dfrac{c^2-d^2}{c-d}
= -\dfrac{(c-d)\cdot (c+d) }{c-d}
= - (c+d)
$}}\\\\$$

heureka  Mar 11, 2015
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2+0 Answers

 #3
avatar+18845 
+10
Best Answer

$$\small{\text{
$
\dfrac{
\dfrac{c}{d}-\dfrac{d}{c}
}
{
\dfrac{1}{c}-\dfrac{1}{d}
}
=
\dfrac{
\dfrac{c^2-d^2}{dc}
}
{
\dfrac{d-c}{dc}
}
= \left( \dfrac{c^2-d^2}{dc} \right) \cdot
\left( \dfrac{dc}{d-c} \right)
= \dfrac{c^2-d^2}{d-c}
= -\dfrac{c^2-d^2}{c-d}
= -\dfrac{(c-d)\cdot (c+d) }{c-d}
= - (c+d)
$}}\\\\$$

heureka  Mar 11, 2015
 #4
avatar
0

Thanks. That's the answer I also got, when I tried it the second time. :)

Guest Mar 11, 2015

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