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Question

If $$a = \dfrac{{4xy}}{{x + y}},$$ the value of $$\dfrac{{a + 2x}}{{a - 2x}} + \dfrac{{a + 2y}}{{a - 2y}}$$ In most simplified form is


A
0
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B
1
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C
1
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D
2
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Solution

The correct option is D $$2$$
Solution : $$ a = \frac{4xy}{x+y} $$
then $$ \frac{a+2x}{a-2x} = \frac{\frac{4xy}{x+y}+2x}{\frac{4xy}{x+y}-2x} = \frac{4xy+2x^{2}+2xy}{4xy-2x^{2}-2xy} = \frac{2x^{3}+6xy}{-2x^{2}+2xy} $$
$$ = \frac{x+3xy}{-x+y} $$ 
And $$ \frac{a+2y}{a-2y} = \frac{\frac{4xy}{x+y}+2y}{\frac{4xy}{x+y}-2y} = \frac{4xy+2xy+2y^{2}}{2xy-2y^{2}} = \frac{6xy+2y^{2}}{2xy-2y^{2}} $$
$$ = \frac{3x+2y}{x-y} $$
then $$\frac{a+2x}{a-2x}+\frac{a+2y}{a-2y} = \frac{x+3y}{-x+y}+\frac{3x+2y}{x-y} $$
$$= \frac{x^{2}-xy+3xy-3y^{2}-3x^{2}-2xy+3xy+y^{2}}{-x^{2}+2xy-y^{2}} $$
$$= \frac{-2x^{2}-2y^{2}+4xy}{-x^{2}+2xy-y^{2}} = \frac{-2[x^{2}+y^{2}-2xy]}{-[x^{2}+y^{2}-2xy]} $$
$$ = 2 $$ 

1123059_1183100_ans_7070898914a4414ab47e847069271046.jpg

Mathematics

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