+0  
 
0
1215
2
avatar

$${\frac{\left({\frac{{\mathtt{1}}}{{{\mathtt{x}}}^{{\mathtt{2}}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{{\mathtt{y}}}^{{\mathtt{2}}}}}\right)}{\left({\frac{{\mathtt{1}}}{{\mathtt{x}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{y}}}}\right)}}$$ =?

 Nov 20, 2014

Best Answer 

 #2
avatar+26400 
+10

$${\frac{\left({\frac{{\mathtt{1}}}{{{\mathtt{x}}}^{{\mathtt{2}}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{{\mathtt{y}}}^{{\mathtt{2}}}}}\right)}{\left({\frac{{\mathtt{1}}}{{\mathtt{x}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{y}}}}\right)}}$$ = ?

$$\dfrac{
\left( \dfrac{1}{x^2} + \dfrac{1}{y^2} \right)
}
{
\left( \dfrac{1}{x} + \dfrac{1}{y} \right)
}
=
\dfrac{
\left( \dfrac{1}{x} + \dfrac{1}{y} \right)\left( \dfrac{1}{x} + \dfrac{1}{y} \right) -2\times\dfrac{1}{x}\times\dfrac{1}{y}
}
{
\left( \dfrac{1}{x} + \dfrac{1}{y} \right)
}\\ \\ \\
=
\dfrac{
\left( \dfrac{1}{x} + \dfrac{1}{y} \right)
\left( \dfrac{1}{x} + \dfrac{1}{y} \right)
}
{
\left( \dfrac{1}{x} + \dfrac{1}{y} \right)
}-
\dfrac{
\dfrac{2}{xy}}
{
\left( \dfrac{1}{x} + \dfrac{1}{y} \right)
}
\\ \\ \\
=
\left( \dfrac{1}{x} + \dfrac{1}{y} \right)
-
\dfrac{2}{xy\left( \dfrac{1}{x} + \dfrac{1}{y} \right)
}
\\ \\ \\
=
\left( \dfrac{1}{x} + \dfrac{1}{y} \right)
-
\dfrac{2}{x+y}$$

.
 Nov 20, 2014
 #1
avatar+23254 
+5

[ 1/x² + 1/y² ] / [ 1/x + 1/y ]

Look at all the denominators; their common denominator is x²y².

Multiply both the numerator and the denominator of the fraction by x²y²:

Numerator:  [ 1/x² + 1/y² ] · x²y²  =  y² + x²  =  x² + y²

Denominator:  [ 1/x + 1/y ] · x²y²  =  xy² + x²y  =  xy(x + y)

Answer:  [  x² + y² ] / [ xy(x + y) ]

(You can't simplify any farther.)

 Nov 20, 2014
 #2
avatar+26400 
+10
Best Answer

$${\frac{\left({\frac{{\mathtt{1}}}{{{\mathtt{x}}}^{{\mathtt{2}}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{{\mathtt{y}}}^{{\mathtt{2}}}}}\right)}{\left({\frac{{\mathtt{1}}}{{\mathtt{x}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{y}}}}\right)}}$$ = ?

$$\dfrac{
\left( \dfrac{1}{x^2} + \dfrac{1}{y^2} \right)
}
{
\left( \dfrac{1}{x} + \dfrac{1}{y} \right)
}
=
\dfrac{
\left( \dfrac{1}{x} + \dfrac{1}{y} \right)\left( \dfrac{1}{x} + \dfrac{1}{y} \right) -2\times\dfrac{1}{x}\times\dfrac{1}{y}
}
{
\left( \dfrac{1}{x} + \dfrac{1}{y} \right)
}\\ \\ \\
=
\dfrac{
\left( \dfrac{1}{x} + \dfrac{1}{y} \right)
\left( \dfrac{1}{x} + \dfrac{1}{y} \right)
}
{
\left( \dfrac{1}{x} + \dfrac{1}{y} \right)
}-
\dfrac{
\dfrac{2}{xy}}
{
\left( \dfrac{1}{x} + \dfrac{1}{y} \right)
}
\\ \\ \\
=
\left( \dfrac{1}{x} + \dfrac{1}{y} \right)
-
\dfrac{2}{xy\left( \dfrac{1}{x} + \dfrac{1}{y} \right)
}
\\ \\ \\
=
\left( \dfrac{1}{x} + \dfrac{1}{y} \right)
-
\dfrac{2}{x+y}$$

heureka Nov 20, 2014

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