Question

|xy-(x^2+y^2)/2|+|(x^2y^2)/2+xy|-2y^2​

Answer 1

$\left| xy - \dfrac{x^{2} + y^{2}}{2} \right| + \left| \dfrac{x^{2} + y^{2}}{2} + xy \right| - 2y^{2}$

$1) \ \left| xy - \dfrac{x^{2} + y^{2}}{2} \right|$

По свойству модуля $|a| =|-a|$ имеем:

$\left| xy - \dfrac{x^{2} + y^{2}}{2} \right| = \left| -\left(\dfrac{x^{2} + y^{2}}{2} - xy \right) \right| = \left| \dfrac{x^{2} + y^{2}}{2} - xy \right|$

Приведем к общему знаменателю:

$\left| \dfrac{x^{2} + y^{2}}{2} - xy \right| = \left| \dfrac{x^{2} - 2xy + y^{2}}{2} \right| = \left| \dfrac{(x - y)^{2}}{2} \right|$

Имеем неотрицательное подмодульное выражение.

По свойству модуля $|a| = a,$ если $a \geqslant 0,$ имеем:

$\left| \dfrac{(x - y)^{2}}{2} \right| = \dfrac{(x - y)^{2}}{2}$

Аналогично:

$2) \ \left| \dfrac{x^{2} + y^{2}}{2} + xy \right| = \dfrac{(x + y)^{2}}{2}$

Тогда имеем:

$\dfrac{(x - y)^{2}}{2} + \dfrac{(x + y)^{2}}{2} - 2y^{2}$

$\dfrac{x^{2} - 2xy + y^{2}}{2} + \dfrac{x^{2} + 2xy + y^{2}}{2} - 2y^{2}$

$\dfrac{x^{2} - 2xy + y^{2} + x^{2} + 2xy + y^{2} - 4y^{2}}{2}$

$\dfrac{2x^{2} - 2y^{2}}{2}$

$\dfrac{2(x^{2} - y^{2})}{2}$

$x^{2} - y^{2}$

Ответ: $x^{2} - y^{2}$

Answer 2

It seems like you've provided an expression involving variables x and y:

|xy - (x^2 + y^2)/2| + |(x^2 * y^2)/2 + xy| - 2y^2

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