Question

$\text{The value of the ratio}\frac{4xy}{(x+y{)}^2-(x-y{)}^2}$
$\text{is equal to}$

• A. 0
• B. 1
• C. -1
• D. 2

Answer 1

The correct option is B 1

$\text{The ratio is given as:}\frac{4xy}{(x+y{)}^2-(x-y{)}^2}$

Now, you know that ${(x+y)}^2=x^2+2xy+y^2$
and ${(x-y)}^2=x^2-2xy+y^2$.

If you put the respective values in the denominator, you will get:
$(x+y{)}^2-(x-y{)}^2$
$=(x{)}^2+2xy+(y{)}^2-\left(\right(x{)}^2-2xy+\left(y{)}^2\right)$
$=x^2+2xy+y^2-(x^2-2xy+y^2)$
$=x^2+2xy+y^2-x^2+2xy-y^2$
$=4xy$

$\text{∴ The ratio will become =}\frac{4xy}{4xy}=1$