Question

√x2-4Х-1=2
x-√x+1=5
√х√2-х=2х
х-1 =√-1
√2х-1

Answer 1

Пошаговое объяснение:

1)

$\sqrt{ {x}^{2} - 4x - 1 } = 2 \\ {x}^{2} - 4x - 1 = 4 \\ {x}^{2} - 4x - 5 = 0 \\ x1 = 5 \\ x2 = - 1$

2)

$x - \sqrt{x + 1} = 5 \\ \sqrt{x + 1} =x - 5 \\ x + 1 = {x }^{2} - 10x + 25 \\ x \geqslant 5 \\ \\ {x}^{2} - 11x + 24 = 0 \\ x1 = 8 \\ x2 = 3$

x2 - не подходит, так как < 5

$\sqrt{x} \sqrt{2 - x} = 2x \\ x(2 - x) = 4 {x}^{2} \\ 0 \leqslant x \leqslant 2 \\ \\2x - {x}^{2} = 4 {x}^{2} \\ 5 {x}^{2} - 2x = 0 \\ x(5x - 2) = 0 \\ x = 0 \\ x = 0.4$