smurfy14
  • smurfy14
[(1/(x^2)-1/Y^2)]/[(1/x^2+2/xy+1/y^2)] plz help! youd be awsome if ya did! (tell me how you get then answer as well, thanks!)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
What are your instructions for this problem?
smurfy14
  • smurfy14
"simplify complex functions"
anonymous
  • anonymous
OK, I started by multiplying the top and bottom of the complex fraction by the LCD, which is x^2y^2 in this case

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More answers

anonymous
  • anonymous
\[\frac{\frac{1}{x^2}-\frac{1}{2}}{\frac{1}{x^2}+\frac{2}{xy}+\frac{1}{y^2}}\]
anonymous
  • anonymous
arya has it, just wanted to type it.
anonymous
  • anonymous
Which came out to (y^2-x^2)/(y^2+2xy+x^2)
anonymous
  • anonymous
Sorry, haven't figured out how to type all of the nice equations yet
anonymous
  • anonymous
there is a gimmick here. the denominator is \[(\frac{1}{x}+\frac{1}{y})^2\]
anonymous
  • anonymous
and the numerator is \[(\frac{1}{x}+\frac{1}{y})(\frac{1}{x}-\frac{1}{y})\]
anonymous
  • anonymous
so you can cancel.
smurfy14
  • smurfy14
can you explain how multiplying the LCD with the second fraction on the bottom came out to be 2xy?
smurfy14
  • smurfy14
nevermind i got it
anonymous
  • anonymous
\[(2/xy)timesx ^{2}y ^{2}=2x ^{2}y ^{2}/xy=2\]
smurfy14
  • smurfy14
idk what to do after (y^2-x^2)/(y^2+2xy+x^2)
anonymous
  • anonymous
Ignore my response, it didn't come out quite right
anonymous
  • anonymous
Factor
anonymous
  • anonymous
\[\frac{a^2-b^2}{a+2ab+b^2}=\frac{(a+b)(a-b)}{(a+b)(a+b)}=\frac{a-b}{a+b}\]
anonymous
  • anonymous
y^2-x^2=(y-x)(y+x) y^2+2xy+x^2=(y+x)(y+x)
anonymous
  • anonymous
\[\frac{\frac{1}{x}-\frac{1}{y}}{\frac{1}{x}+\frac{1}{y}}\] multiply top an bottom by \[xy\]
anonymous
  • anonymous
Then we can cancel one of the (y+x)
smurfy14
  • smurfy14
oh ok thanks so much! think you could help me with [a-b]/[a^-1=b^-1]
anonymous
  • anonymous
Sure
smurfy14
  • smurfy14
imean [a-b]/[a^-1-b^-1]
anonymous
  • anonymous
OK, a^-1=1/a and b^-1=1/b
anonymous
  • anonymous
Does that part make sense?
smurfy14
  • smurfy14
no? so ur saying you just move the -1 to the top and not the a along with it?
anonymous
  • anonymous
Not quite, lets see if it makes more sense when I type out the whole equation. Hang on
anonymous
  • anonymous
(a-b)/(a^-1-b^-1) = (a-b)/[(1/a)-(1/b)]
anonymous
  • anonymous
Does that make any more sense or no?
smurfy14
  • smurfy14
oh ok ya that makes sense, so you dont move the a with the 1 im guessing
anonymous
  • anonymous
\[\frac{\frac{1}{x^2}-\frac{1}{y^2}}{\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{x y}}=1-\frac{2 x}{x+y} \] Both equation sides evaluated at x=11 and y=17 yield the value 3/14. Note: The first "y" from left to right in the problem expression text string is a cap y, Y, not lower case y. Mathematica views the two as different characters. The cap Y was changed to lower case prior to solving the problem.
anonymous
  • anonymous
Not quite sure what you mean
smurfy14
  • smurfy14
hah nevermind i got it you can keep going :)
anonymous
  • anonymous
Cool, so then just like the previous problem the next step is to multiply the to and bottom by the LCD, which in this case is ab
anonymous
  • anonymous
*top
smurfy14
  • smurfy14
so (ab-ab)/(0)?
smurfy14
  • smurfy14
how do you figure out the denominator?
anonymous
  • anonymous
I got (a-b)ab for the numerator and b-a for the denominator
anonymous
  • anonymous
To get the denominator: (1/a-1/b)ab (the LCD)=ab/a-ab/b=b-a
smurfy14
  • smurfy14
ok got it so how did you get the numerator?
anonymous
  • anonymous
I just multiplied (a-b) by the LCD: ab and got (a-b)ab
smurfy14
  • smurfy14
oh duh lol so would you simplify that farther?
anonymous
  • anonymous
Yes, at this point the equation looks like [(a-b)ab]/(b-a) and if we factor out -1 from the denominator we get [(a-b)ab]/[-(a-b)]
anonymous
  • anonymous
Then the (a-b) cancels and we get -ab
smurfy14
  • smurfy14
oh ok thank you soo much!!
anonymous
  • anonymous
No problem, glad I could help! :)
anonymous
  • anonymous
\[\frac{\frac{1}{x^2}-\frac{1}{y^2}}{\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{x y}}=\frac{\frac{-x^2+y^2}{x^2 y^2}}{\frac{x^2+2 x y+y^2}{x^2 y^2}}=\frac{-x^2+y^2}{x^2+2 x y+y^2}=\frac{(-(x-y) )}{(x+y)}=\frac{-x+y}{x+y}\]\[1-\frac{2 x}{x+y}=\frac{-x+y}{x+y} \]

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