Question

if xy=1 what is the value of: 2(x+y)^2/2 (x-y)^2

Answer 1

so to solve this exercise you need to know some formules

(a+b)^2 = ?

(a-b)^2 = ?

do you know ?

Answer 2

(a+b)^2 = (a+b)(a+b) = a2+ 2ab + b2
(a-b)^2 = (a-b)(a-b) = a2- 2ab + b2

Answer 3

so than now you just rewrite it using a=x and b=y

ok ?

Answer 4

yes but if x*y = 1 how do i know the values of x and y
it could be 1 * 1 or 2 * 0,5

Answer 5

no

so you know that x*y = 1 from what you can getting that x = 1/y

using this you need just substitute inside this equation

Answer 6

2(x+y)^2/2 (x-y)^2

x = 1/y

2(1/y +y)^2 /

so the denominator is 2(x-y)^2

???

Answer 7

how is right please ?

Answer 8

no idea i'm bad at math xD
There are 5 mc anwsers: 1, 2, 4, 16 and 19

If i take x=1 and y=1 it's 8/0 which is impossible
if x=2 and y=1/2 the answer is 2,777.....
if x=3 and y=1/3 the answer is 1,546.....

if x=1/2 and y =2 the answer is 9
if x=1/3 and y=3 the answer is 4

but how can i ever know that x=1/3 and y=3 is the right combination

Answer 9

I think the equation you wrote is missing an "=" sign....
could you rewrite it using the "equation" button below so we understand correctly

Answer 10

already solved it:

\[2(x+y)^2/2(x-y)^2 \]\[= 2((x+y)^2 - (x-y)^2) \]\[= 2^{4xy}\]\[= 2^{4*1}\]\[= 16\]

Answer 11

There is no unique value for
\[ \frac{2(x+y)^2}{2 (x-y)^2} \]

do you mean
\[ 2(x+y)^2 - 2(x-y)^2 \]?

Answer 12

For the 2nd case, expand each square (use FOIL) Example: (x+y)^2= x^2 + 2xy + y^2
and simplify

Answer 13

lol i'll just stick with biology and chemistry xD
i hate this kind of math :')