Question

Решить уравнения 1)3(x+1)=18. 2)(y-5)(y-8)=y^2+1. 3)(3x+2)(4x-1)-(6x+2)(2x-1)=0

Answer 1

3(х + 1) = 18
3х + 3 = 18
3х = 18 - 3
3х = 15
  х = 5
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(у - 5)(у - 8) = у^2 + 1
y^2 - 5y - 8y + 40 = y^2 + 1
- 13y + 40 = 1
- 13y = - 40 + 1
- 13y = - 39
      y = 3
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(3x + 2)(4x - 1) - (6x + 2)(2x - 1) = 0
(12x^2 + 8x - 3x - 2 ) - (12x^2 + 4x - 6x - 2) = 0
12x^2 + 5x - 2 - 12x^2 + 2x + 2 = 0
7x = 0
  x = 0
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Answer 2

1) 3(х+1)=18
3х+3=18
3х=18-3
3х=15
х=5

3) (3х+2)(4х-1)-(6х+2)(2х-1)=0
12х(х во второй степени)-3х+8х-2-12х(х во 2 степени)+6х+4х-2=0
-3х+8х+6х+4х=2+2
15х=4
х=0,26