Two numbers x and y (x > y) such that their sum is equal to three times their difference. Then value of \( \Large \frac{3xy}{2 \left(x^{2}-y^{2}\right) } \) will be;
Correct Answer: Description for Correct answer:
x > y
given :-
\( \Large x+y = 3 \left(x-y\right) \)
=> \( \Large x + y = 3x - 3y \)
=> \( \Large x - 3x = -3y - y \)
=> \( \Large -2x = -4y => x = 2y \)
Therefore, \( \Large \frac{3xy}{2 \left(x^{2}-y^{2}\right) }=\frac{3 \times 2y \times y}{2 \left( \left(2y\right)^{2}-y^{2} \right) } \)
= \( \Large \frac{6y^{2}}{2 \times \left(4y^{2}-y^{2}\right) } \)
= \( \Large \frac{6y^{2}}{6y^{2}} = 1 \)
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