Замените звездочки такими одночленами, чтобы образовалось тождество:
1) $(x - y) ⋅ * = x^2y^2 - x^3y$;
2) $(-9x^2 + *) ⋅ y = * + y^4$;
3) $(1,4x - *) ⋅ 3x = * - 0,6x^3$;
4) $*(* - x^2y^5 + 5y^6) = 8x^3y^3 + 5x^3y^8 - *$.
$(x - y) ⋅ * = x^2y^2 - x^3y$
$(x - y) ⋅ * = -x^2y(x - y)$
$* = -\frac{x^2y(x - y)}{x - y}$
$* = -x^2y$
$(-9x^2 + *_1) ⋅ y = *_2 + y^4$
$(-9x^2 + *_1) = \frac{*_2 + y^4}{y}$
$(-9x^2 + *_1) = \frac{-9x^2y + y^4}{y}$
$*_2 = -9x^2 * y$
$*_2 = -9x^2y$;
$(-9x^2 + *_1) ⋅ y = -9x^2y + y^4$
$(-9x^2 + *_1) = \frac{-9x^2y + y^4}{y}$
$-9x^2 + *_1 = -9x^2 + y^3$
$*_1 = -9x^2 + y^3 + 9x^2$
$*_1 = y^3$
$(1,4x - *_1) ⋅ 3x = *_2 - 0,6x^3$
$(1,4x - *_1) = \frac{*_2 - 0,6x^3}{3x}$
$*_2 = 1,4x * 3x$
$*_2 = 4,2x^2$;
$(1,4x - *_1) ⋅ 3x = 4,2x^2 - 0,6x^3$
$(1,4x - *_1) = \frac{4,2x^2 - 0,6x^3}{3x}$
$1,4x - *_1 = 1,4x - 0,2x^2$
$-*_1 = 1,4x - 0,2x^2 - 1,4x$
$-*_1 = -0,2x^2$
$*_1 = 0,2x^2$
$*_1(*_2 - x^2y^5 + 5y^6) = 8x^3y^3 + 5x^3y^8 - *_3$
$*_1 = -5xy^3$, $*_2 = -1,6x^2$, $*_3 = 25xy^3$:
$-5xy^3(-1,6x^2 - x^2y^5 + 5y^6) = 8x^3y^3 + 5x^3y^8 - 25xy^9$
$8x^3y^3 + 5x^3y^8 - 25xy^9 = 8x^3y^3 + 5x^3y^8 - 25xy^9$
Пожауйста, оцените решение
