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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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Math Expert
Joined: 02 Sep 2009
Posts: 60647
If x and y are non-zero numbers, adding the reciprocals of x^2 and y^2  [#permalink]

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05 Apr 2016, 08:40
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Difficulty:

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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Joined: 02 Aug 2009
Posts: 8336
Re: If x and y are non-zero numbers, adding the reciprocals of x^2 and y^2  [#permalink]

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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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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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Sortem sternit fortem!
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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