+0  
 
0
892
2
avatar

Simplify. Is this correct?

My answer = c^2 + d^2

 Mar 10, 2015

Best Answer 

 #3
avatar+26367 
+10

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

.
 Mar 11, 2015
 #3
avatar+26367 
+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. :)

 Mar 11, 2015

1 Online Users