Question

Помогите решить отмеченные номера по алгебре. Пожалуйста))

Answer 1

2(a). D = 36 - 36 = 0
x = -6/2 = -3
(x+3)²
2(b).
D = 4 + 32 = 6²
x1 = 5
x2 = -2
(x+2)(x-5)
3(a). x²-4 = (x-2)(x+2)
x²-5x+6 = (x-2)(x-3)
D = 25 - 24 = 1
x1 = 3
x2 = 2
(x-2)(x+2)/(x-2)(x-3) = (x+2)/(x-3)

Answer 2

лист I.
1. 
a)
х² - 5х = 0
х (х - 5) = 0
х₁ = 0
х - 5 =0
х₂ = 5
б)
4а² +3а =0
а(4а+3) =0
а₁ = 0
4а+3=0
4а=3
а= 3/4
а₂= 0,75
2.
а)
(а+3)(а-2) = -6
а*а+а*(-2) +3*а +3*(-2) = -6
а²  -2а + 3а  - 6 = -6
а²  + а  = - 6 +6
а² + а=0
а(а+1) = 0
а₁ =0
а+1=0
а₂ = -1

лист II.
2.
а) у²  +6у  + 9 = у²  + 2*у*3 + 3²  =(у+3)² = (у+3)(у+3)
б) х²  - 2х -  8 = x² + 2*x*(-1)+(-1)²   -  9 = (x - 1)²  - 3² =
= (x-1-3)(x-1+3) = (x - 4)(x + 2)

3.
$a) \frac{x^2 - 4}{x^2 -5x + 6} = \frac{x^2 -2^2}{x^2 - 2x -3x + 6} = \frac{(x-2)(x+2)}{x(x-2) -3(x -2)} = \frac{(x-2)(x+2)}{(x-2)(x-3)} = \frac{x+2}{x-3} \\ \\ b) \frac{5x-x^2}{x^2 +x - 30} = \frac{-x(x-5)}{x^2 -5x + 6x - 30} = \frac{-x(x-5)}{x(x-5)+6(x-5)} = \frac{-x(x-5)}{(x+6)(x-5)} = \frac{-x}{x+6} = - \frac{x}{x+6} \\ \\ c) \frac{a^2 -2a+1}{2a^2 -3a+1} = \frac{(a-1)^2}{2a^2-2a-a+1} =\frac{(a-1)(a-1)}{2a(a-1) -1(a-1)} = \frac{(a-1)(a-1)}{(2a-1)(a-1)} =\frac{a-1}{2a-1}$