Алгебра 9 класс учебник Макарычев, Миндюк ответы – номер 479

пусть x + y = u, xy = v

x² + y² = x² + y² + 2xy – 2xy = (x + y)² – 2xy = u² – 2v

u² – 5 + u – 7 = 0
u² + u – 12 = 0

По теореме Виета
u₁ = 3, u₂ = –4
v₁ = 5 – 3 = 2
v₂ = 5 – (–4) = 9

3y – y² – 2 = 0
y² – 3y + 2 = 0

По теореме Виета
y₁ = 1, y₂ = 2
x₁ = 3 – 1 = 2
x₂ = 3 – 2 = 1

–4y – y² – 9 = 0
y² + 4y + 9 = 0
D = 4² – 4 · 1 · 9 = 16 – 36 = –20 < 0
корней нет

Ответ: (2; 1), (1; 2)

пусть x + y = u, xy = v

x² + y² = x² + y² + 2xy – 2xy = (x + y)² – 2xy = u² – 2v

u² – 1 + u – 19 = 0
u² + u – 20 = 0

По теореме Виета
u₁ = –5, u₂ = 4
v₁ = 1 – (–5) = 6
v₂ = 1 – 4 = –3

–5y – y² – 6 = 0
y² + 5y + 6 = 0

По теореме Виета
y₁ = –2, y₂ = –3
x₁ = –5 – (–2) = –3
x₂ = –5 – (–3) = –2

4y – y² + 3 = 0
y² – 4y – 3 = 0
D = (–4)² – 4 · 1 · 3 = 28

у = $$\frac{4 ± \sqrt{28}}{2 \cdot 1}$$ = $$\frac{4 ± 2\sqrt{7}}{2}$$ = 2 ± $$\sqrt{7}$$

у₁ = 2 + $$\sqrt{7}$$
у₂ = 2 – $$\sqrt{7}$$

х₁ = 4 – (2 + $$\sqrt{7}$$) = 4 – 2 – $$\sqrt{7}$$ = 2 – $$\sqrt{7}$$

х₂ = 4 – (2 – $$\sqrt{7}$$) = 4 – 2 + $$\sqrt{7}$$ = 2 + $$\sqrt{7}$$

Ответ: (–3; –2), (–2; –3), (2 –$$\sqrt{7}$$; 2 +$$\sqrt{7}$$),(2 + $$\sqrt{7}$$; 2 – $$\sqrt{7}$$)