Question

Помогите с алгеброй пж​

Answer 1

Объяснение:

1) $y=\sqrt{1-3x}$

D(y): 1-3x >= 0

3x <= 1

x <= 1/3

D(y) = (-∞;1/3]

2) $y = \frac{\sqrt{x-3} }{x+5}$

D(y): x-3 >= 0 ∩ x + 5 ≠ 0

D(y): [3; +∞)

3)  $y = \sqrt{2-x} + \sqrt{3x+1}$

D(y): 2-x >= 0 ∩ 3x+1 >= 0

2-x >= 0   ->  x <= 2

3x+1 >= 0   ->  x >= -1/3

D(y) = [-1/3; 2]

4) $y = \sqrt{x-3} + \sqrt{3-x}$

D(y): x-3 >= 0 ∩ 3-x >= 0

x-3 >= 0   ->   x >= 3

3-x >= 0   ->   x <= 3

D(y) = {3}