Question

Please help in the math


`(x + y)/(x - y) + (x - y)/(x + y) - \frac{2( {x}^(2) - {y}^(2)) }{ {x}^(2) - {y}^(2) }`
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Answer 1

9514 1404 393

Answer:

4y²/(x² -y²)

Explanation:

The expression simplifies as follows:

`(x+y)/(x-y)+(x-y)/(x+y)-(2(x^2-y^2))/(x^2-y^2)\\\\=((x+y)(x+y)+(x-y)(x-y)-2(x^2-y^2))/((x-y)(x+y))\\\\=((x+y)^2+(x-y)^2-2(x^2-y^2))/(x^2-y^2)\\\\=((x^2+2xy+y^2)+(x^2-2xy+y^2)-2(x^2-y^2))/(x^2-y^2)\\\\=(2(x^2+y^2-(x^2-y^2)))/(x^2-y^2)=\boxed{(4y^2)/(x^2-y^2)}`

Answer 2

`\rm \displaystyle (x + y)/(x - y) + (x - y)/(x + y) - \frac{2( {x}^(2) - {y}^(2)) }{ {x}^(2) - {y}^(2) } = \boxed{ \displaystyle (4y ^2)/((x - y)(x + y)) }`

Explanation:

we want to simplify the following

`\rm \displaystyle (x + y)/(x - y) + (x - y)/(x + y) - \frac{2( {x}^(2) - {y}^(2)) }{ {x}^(2) - {y}^(2) }`

notice that we can reduce the fraction thus do so:

`\rm \displaystyle (x + y)/(x - y) + (x - y)/(x + y) - \frac{2 \cancel{( {x}^(2) - {y}^(2)) }}{ \cancel{{x}^(2) - {y}^(2) }}`

`\rm \displaystyle (x + y)/(x - y) + (x - y)/(x + y) - 2`

in order to simplify the addition of the algebraic fraction the first step is to figure out the LCM of the denominator and that is (x-y)(x+y) now divide the LCM by the denominator of very fraction and multiply the result by the numerator which yields:

`\rm \displaystyle (x + y)/(x - y) + (x - y)/(x + y) - 2 \\ \\ \displaystyle ((x + y)^2 + (x - y)^2 - 2(x + y)(x - y))/((x - y)(x + y))`

factor using (a-b)²=a²+b²-2ab

`\rm \displaystyle ((x + y-(x - y) )^2)/((x - y)(x + y))`

remove parentheses

`\rm \displaystyle ((x + y-x + y) )^2)/((x - y)(x + y))`

simplify:

`\rm \displaystyle (4y ^2)/((x - y)(x + y))`