Алгебра 9 класс учебник Макарычев, Миндюк ответы – номер 498

1 · (2x – 5) · 6 + 1 · 6x = 1 · x(2x – 5)

12x – 30 + 6x = 2x² – 5x

2x² – 5x – 18x + 30 = 0

2x² – 23x + 30 = 0

D = (–23)² – 4 · 2 · 30 = 529 – 240 = 289

х = $$\frac{23 ± \sqrt{289}}{2 \cdot 2}$$ = $$\frac{23 ± \sqrt{17}}{4}$$

x₁ = 10, x₂ = 1,5
y₁ = 2 · 10 – 5 = 15
y₂ = 2 · 1,5 – 5 = –2

Ответ: (10; 15), (1,5; –2)

1 · y · 20 – 1 · 20(14 – 2y) = 1 · y(14 – 2y)

20y – 280 + 40y = 14y – 2y²

2y² – 14y + 60y – 280 = 0

2y² + 46y – 280 = 0

y² + 23y – 140 = 0

D = 23² – 4 · 1 · (–140) = 529 + 560 = 1089

у = $$\frac{-23 ± \sqrt{1089}}{2 \cdot 1}$$ = $$\frac{-23 ± \sqrt{33}}{2}$$

y₁ = – 28, y₂ = 5
x₁ = 14 – 2 · (–28) = 70
x₂ = 14 – 2 · 5 = 4

Ответ: (70; –28), (4; 5)

12 · (14 – y)(14 – y) + 12 · y · y = 25 · y(14 – y)

12 · (196 – 28y + y²) + 12y² = 350y – 25y²

2352 – 336y + 12y² + 12y² – 350y + 25y² = 0

49y² – 686y + 2352 = 0

y² – 14y + 48 = 0

По теореме Виета
y₁ = 6, y₂ = 8
x₁ = 14 – 6 = 8
x₂ = 14 – 8 = 6

Ответ: (8; 6), (6; 8)

6 · (2 + y)(2 + y) – 6 · y · y = 5 · y(2 + y)

6 · (4 + 4y + y²) – 6y² = 10y + 5y²

24 + 24y + 6y² – 6y² – 10y – 5y² = 0

–5y² + 14y + 24 = 0

5y² – 14y – 24 = 0

D₁ = (–7)² – 5 · (–24) = 49 + 120 = 169

у = $$\frac{7 ± \sqrt{169}}{5}$$ = $$\frac{7 ± \sqrt{13}}{5}$$

y₁ = 4, y₂ = –1,2
x₁ = 2 + 4 = 6
x₂ = 2 + (–1,2) = 0,8

Ответ: (6; 4), (0,8; –1,2)