Алгебра 9 класс учебник Макарычев, Миндюк ответы – номер 492

1 + 6y + 9y² – 2y – 6y² + y – 3 = 0
3y² + 5y – 2 = 0
D = 5² – 4 · 3 · (–2) = 25 + 24 = 49

у = $$\frac{-5 ± \sqrt{49}}{2 \cdot 3}$$ = $$\frac{-5 ± \sqrt{7}}{6}$$
у₁ = –2, у₂ = 2/6 = 1/3
х₁ = –1 – 3 · (–2) = 5
х₂ = –1 – 3 · 1/3 = –2

Ответ: (5; –2), (–2; 1/3)

2x² – x – 4x² + 4x – 1 + 3x + 1 = 0
–2x² + 6x = 0
–2x (x – 3) = 0

–2x = 0  или  х – 3 = 0
x₁ = 0             x₂ = 3

y₁ = 2 · 0 – 1 = – 1
y₂ = 2 · 3 – 1 = 5

Ответ: (0; – 1), (3; 5)

y – 5y – 11 + y² + 6 = 0
y² – 4y – 5 = 0

По теореме Виета
y₁ = – 1, y₂ = 5

2x – 1 – 11 = 0
2x = 12
x₁ = 6
2x + 5 – 11 = 0
2x = 6
x₂ = 3

Ответ: (6; –1), (3; 5)

2y² + 16y + 32 – 3y² – 5y – 20 – 2y – 26 = 0
–y² + 9y – 14 = 0
y² – 9y + 14 = 0

По теореме Виета
y₁ = 2, y₂ = 7

x₁ = 2 + 4 = 6
x₂ = 7 + 4 = 11

Ответ: (6; 2), (11; 7)

4(2,25y² + 7,5y + 6,25) – 9y² – 38,5y + 2,5 – 19 = 0
9y² + 30y + 25 – 9y² – 38,5y – 16,5 = 0
–8,5y + 8,5 = 0
–8,5y = –8,5
y = 1
x = 1,5 · 1 + 2,5 = 4

Ответ: (4; 1)

3y² – 12y + 12 + y² + 8y – 16 + 13y – 5 = 0
4y² + 9y – 9 = 0
D = 9² – 4 · 4 · (–9) = 81 + 144 = 225

у = $$\frac{-9 ± \sqrt{225}}{2 \cdot 4}$$ = $$\frac{-9 ± \sqrt{15}}{8}$$

y₁ = –3, y₂ = 6/8 = 3/4
x₁1 = –3 – 2 = –5
x₂ = 3/4 – 2 = –11/4

Ответ: (–5; –3), (–11/4; 3/4)