Алгебра 9 класс учебник Макарычев, Миндюк ответы – номер 473

2y² = 8
y² = 4
y₁ = 2, y₂ = −2
x₁ = 2, x₂ = −2

1 – 4y + 4y² + y² = 8
5y² – 4y – 7 = 0
D₁ = (–2)² – 5 · (–7) = 4 + 35 =39

у = $$\frac{2 ± \sqrt{39}}{5}$$

у₁ = $$\frac{2 + \sqrt{39}}{5}$$, у₂ = $$\frac{2 - \sqrt{39}}{5}$$

х₁ = 1 – 2 · $$\frac{2 + \sqrt{39}}{5}$$ = $$\frac{5 - 4 - 2\sqrt{39}}{5}$$ = $$\frac{1 - 2\sqrt{39}}{5}$$

х₂ = 1 – 2 · $$\frac{2 - \sqrt{39}}{5}$$ = $$\frac{5 - 4 + 2\sqrt{39}}{5}$$ = $$\frac{1 + 2\sqrt{39}}{5}$$

Ответ: (–2; 2), ($$\frac{1 - 2\sqrt{39}}{5}$$; $$\frac{2 + \sqrt{39}}{5}$$),($$\frac{1 + 2\sqrt{39}}{5}$$; $$\frac{2 - \sqrt{39}}{5}$$)

y² + 2y + 1 – 20y² – 5 = 0
–19y² + 2y – 4 = 0
19y² – 2y + 4 = 0
D₁ = (–1)² – 19 · 4 = 1 – 76 = –75 < 0
корней нет

25y² – 20y² = 5
5y² = 5
y² = 1
y₁ = 1, y₂ = – 1
x₁ = 5 · 1 = 5
x₂ = 5 · (–1) = – 5

Ответ: (5; 1), (– 5; – 1)