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If x and y are non-zero numbers, adding the reciprocals of x^2 and y^2

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Posts: 60647
If x and y are non-zero numbers, adding the reciprocals of x^2 and y^2  [#permalink]

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New post 05 Apr 2016, 08:40
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E

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  45% (medium)

Question Stats:

66% (01:54) correct 34% (02:31) wrong based on 89 sessions

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If x and y are non-zero numbers, adding the reciprocals of x^2 and y^2 and then subtracting 2 times the reciprocal of xy produce which of the following?

A. \(x^2 + y^2 – 2xy\)

B. \(\frac{(x^2y^2 + xy)}{(x + y^2)}\)

C. \((\frac{1}{y}−\frac{1}{x})\)

D. \(\frac{(x^2 + y^2)}{(x^2y^2)}\)

E. \((\frac{1}{x} − \frac{1}{y})^2\)

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Re: If x and y are non-zero numbers, adding the reciprocals of x^2 and y^2  [#permalink]

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New post 05 Apr 2016, 09:32
Bunuel wrote:
If x and y are non-zero numbers, adding the reciprocals of x^2 and y^2 and then subtracting 2 times the reciprocal of xy produce which of the following?

A. x^2 + y^2 – 2xy
B. (x^2y^2 + xy)/(x + y^2)
C. (1/y−1/x)
D. (x^2 + y^2)/(x^2y^2)
E. (1/x − 1/y)^2



Hi,
The Q may seem complex but in actually is straight forward--
\(\frac{1}{x^2}+\frac{1}{y^2} -\frac{2}{xy}\)

TWO WAYS--


1) substitute
take x = 1/3 and y = 1/2..
\(3^2 +2^2-12\) = \(1\)
substitute values and see..

A. \(x^2 + y^2 – 2xy\)------- 1/9 +1/4 -1/3 -- NO
B. \(\frac{(x^2y^2 + xy)}{(x + y^2)}\)... (1/36 + 1/6)/(1/3+1/4) --NO
C. \((\frac{1}{y}−\frac{1}{x}).\).. 2-3= -1---NO
D. \(\frac{(x^2 + y^2)}{(x^2y^2)-}\)- (1/9+1/4)/(1/36)=13.. NO
E. \((\frac{1}{x}− \frac{1}{y})^2\).. (3-2)^2=1.. CORRECT

2) formula
VERY easy if you can see the hidden formula
\(\frac{1}{x^2}+\frac{1}{y^2} -\frac{2}{xy}\)..
\(\frac{1}{x}^2+\frac{1}{y}^2 - 2*\frac{1}{x}*\frac{1}{y}\)..
\((\frac{1}{x}− \frac{1}{y})^2\)..
same as E
ans E
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Re: If x and y are non-zero numbers, adding the reciprocals of x^2 and y^2  [#permalink]

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New post 27 Jan 2017, 13:05
Bunuel wrote:
If x and y are non-zero numbers, adding the reciprocals of x^2 and y^2 and then subtracting 2 times the reciprocal of xy produce which of the following?

A. \(x^2 + y^2 – 2xy\)

B. \(\frac{(x^2y^2 + xy)}{(x + y^2)}\)

C. \((\frac{1}{y}−\frac{1}{x})\)

D. \(\frac{(x^2 + y^2)}{(x^2y^2)}\)

E. \((\frac{1}{x} − \frac{1}{y})^2\)


Putting x=1, y=2 in \(\frac{1}{x^2} + \frac{1}{y^2} - \frac{2}{xy}\) we get \(\frac{1}{4}\)
Now TEST THE ANSWERS only E fits in.
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Re: If x and y are non-zero numbers, adding the reciprocals of x^2 and y^2   [#permalink] 27 Jan 2017, 13:05
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