please help in the math [tex] \frac{x + y}{x – y} + \frac{x – y}{x + y} – \frac{2( {x}^{2} – {y}^{2}) }{ {x}^{2} – {y}^{2} }

Question

please help in the math
 \frac{x + y}{x - y}  +   \frac{x - y}{x + y}  -   \frac{2( {x}^{2} -  {y}^{2})  }{ {x}^{2}  -  {y}^{2} }

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niczorrrr 3 years 2021-07-18T18:03:55+00:00 2 Answers 7 views 0

Answers ( )

    0
    2021-07-18T18:05:05+00:00

    Answer:

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

    Step-by-step explanation:

    we want to simplify the following

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

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

    remove parentheses

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

    simplify:

     \rm \displaystyle  \frac{4y ^2}{(x - y)(x + y)}

    0
    2021-07-18T18:05:27+00:00

    9514 1404 393

    Answer:

      4y²/(x² -y²)

    Step-by-step explanation:

    The expression simplifies as follows:

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

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )