Алгебра 9 класс учебник Макарычев, Миндюк ответы – номер 854

x³ + (– (x³ + 3x²y + 3xy² + y³)) + y³ = 12
x³ – x³ – 3x²y – 3xy² – y³ + y³ = 12
– 3x²y – 3xy² = 12
x²y + xy² = –4
xy (x + y) = –4
xy · (–xy) = –4
(xy)² = 4
xy = 2 или xy = –2

1) Пусть xy = 2, x = 2/у
2 = –x – y
2/у + y + 2 = 0
2 + y² + 2 = 0
y² + 2y + 2 = 0
D₁ = (–1)² – 1 · 2 = –3 < 0
Корней нет

2) Пусть xy = –2, x = –2/у
2 = –x – y
2/у – y + 2 = 0
2 – y² + 2 = 0
y² – 2y – 2 = 0
D₁ = (–1)² – 1 · (–2) = 3
y = $$\frac{1 ± \sqrt{3}}{1}$$ = 1 ± $$\sqrt{3}$$
y₁ = 1 + $$\sqrt{3}$$
y₂ = 1 – $$\sqrt{3}$$

x₁ = –$$\frac{2}{1 + \sqrt{3}}$$ = $$\frac{2(1 - \sqrt{3})}{(1 + \sqrt{3})(1 - \sqrt{3})}$$ = $$\frac{2(1 - \sqrt{3})}{1 - 3}$$ = $$\frac{2(1 - \sqrt{3})}{-2}$$ = 1 – $$\sqrt{3}$$

x₂ = –$$\frac{2}{1 - \sqrt{3}}$$ = –$$\frac{2(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})}$$ = –$$\frac{2(1 + \sqrt{3})}{1 - 3}$$ = –$$\frac{2(1 + \sqrt{3})}{-2}$$ = 1 + $$\sqrt{3}$$

Ответ: (1 – $$\sqrt{3}$$; 1 + $$\sqrt{3}$$), (1 + $$\sqrt{3}$$; 1 – $$\sqrt{3}$$)