Решите уравнение: а) (6 – х) ⋅ (х + 6) – (х – 11)х = 36; б) (1 – 3y)/11 – (3 – y)/5 = 0; в) 9х² – ((12х – 11)/(3x + 8)) / 4 = 1; г) (y + 1)² / 12 – (1 – y²) / 24 = 4.

Решение:

а) (6 – х) ⋅ (х + 6) – (х – 11)х = 36

6² – x² – (x ⋅ – 11 ⋅ x) – 36 = 0

36 – x² – x² + 11x – 36 = 0

-2x² + 11x = 0

2x² – 11x = 0

x(2x – 11) = 0

x = 0 или 2x – 11 = 0

2x = 11

x = 11 : 2

x = 5,5

x = 0; x = 5,5

б) (1 – 3y)/11 – (3 – y)/5 = 0

5 ⋅ 1 – 3y / 11 – 5 ⋅ 3 – y / 5 = 5 ⋅ 0

5 ⋅ (1 – 3y) – 11 ⋅ (3 – y) = 0

5 ⋅ 1 + 5 ⋅ (-3y) – 11 ⋅ 3 – 11 ⋅ (-y) = 0

5 – 15y – 33 + 11y = 0

-4y – 28 = 0

-4y = 28

y = 28 :(-4)

y = -7

в) 9х² – ((12х – 11)/(3x + 8)) / 4 = 1

9x² – (12x – 11) (3x + 8) / 4 = 1

4 ⋅ 9x² – 4 ⋅ (12x – 11) ⋅ (3x + 8) / 4 = 4 ⋅ 1

36x² – (12x ⋅ 3x + 12x ⋅ 8 – 11 ⋅ 3x – 11 ⋅ 8) = 4

36x² – (36x² + 96x – 33x – 88) = 4

36x² – (36x² + 63x – 88) = 4

36x² – 36x² – 63x + 88 = 4

-63x = -4 – 88

-63x = – 84

x = -84 : (-63)

x = 1 1/3

г) (y + 1)² / 12 – (1 – y²) / 24 = 4

24 ⋅ y² + 2y + 1 / 12 – 24 ⋅ 1 – y² / 24 = 24 ⋅ 4

2(y² + 2y + 1) – (1 – y² ) = 96

2y² + 4y + 2 – 1 + y² – 96 = 0

3y² + 4y – 95 = 0

D = 4² – 4 ⋅ 3 ⋅ (-95) = 16 + 1140 = 1156 больше 0

√D = √1156 = √34² = 34

y₁ = – 4 – 34 / 2 ⋅ 3 = – (-4 + 34) / 6 = – 38/6 = – 6 1/3

y₂ = – 4 + 34 / 2 ⋅ 3 = 34 – 4 / 6 = 30 / 6 = 5

Источник : Учебник по Алгебре 9 класса авторов Ю.Н. Макарычев, Н.Г. Миндюк, К.И. Нешков, С.Б. Суворова, 2014г.