Question

Решите пожалуйста неравенства​

Answer 1

Ответ:

1)

$3x - 2 = {x}^{2} - 4x + 4 \\ {x}^{2} - 4x + 4 - 3x + 2 = 0 \\ {x}^{2} - 7x + 6 = 0 \\ x_{1} = 1 \: \: \: x_{2} = 6$

Подставляем в начало:

$\sqrt{3 \times 1 - 2} = 1 - 2 \\ \sqrt{1} ≠ - 1$

$\sqrt{3 \times 6 - 2} =6 - 2 \\ 4 = 4$

x=6

2)

$x = {x}^{2} - x - 3 \\ {x}^{2} - x - 3 - x = 0 \\ {x}^{2} - 2x - 3 = 0 \\ x_{1} = - 1 \: \: \: \: \: \: \: x_{2} = 3$

$\sqrt{ - 1} ≠ \sqrt{1 + 1 - 3}$

$\sqrt{3} = \sqrt{9 - 3 - 3} \\ \sqrt{3} = \sqrt{3}$

x=3

3)

$2 {x}^{2} - 5x + 1 = {x}^{2} - 2x - 1 \\ 2 {x}^{2} - 5x + 1 - {x}^{2} + 2x + 1 = 0 \\ {x}^{2} - 3x + 2 = 0 \\ x_{1} = 1 \: \: \: \: \: \: \: \: \: \: x_{2} = 2$

$\sqrt{2 \times {1}^{2} - 5 \times 1 + 1} = \sqrt{ {1}^{2} - 2 \times 1 - 1 } \\ \sqrt{ - 2} ≠ \sqrt{ - 2}$

$\sqrt{2 \times {2}^{2} - 5 \times 2 + 1} = \sqrt{ {2}^{2} - 2 \times 2 - 1} \\ \sqrt{ - 1}≠ \sqrt{ - 1}$

$x \in \varnothing$

4)

$x - 5 = 3 - x \\ x - 5 + x - 3 = 0 \\ 2x - 8 = 0 \\ x - 4 = 0 \\ x = 4$

$\sqrt{4 - 5} = \sqrt{3 - 4} \\ \sqrt{ - 1} ≠ \sqrt{ - 1}$

$x \in \varnothing$

5)

$\sqrt{x - 1} = 5 - \sqrt{2x - 1} \\ x - 1 = 25 - 10 \sqrt{2x - 1} + 2x - 1 \\ 10 \sqrt{2x - 1} = 25 + 2x - x \\ 10 \sqrt{2x - 1} = 25 + x \\ 100(2x - 1) = 625 + 50x + {x}^{2} \\ 200x - 100 - 625 - 50x - {x}^{2} = 0 \\ 150x - 725 - {x}^{2} = 0 \\ {x}^{2} - 150x + 725 = 0 \\ x_{1} = 5 \: \: \: \: \: \: \: \: x_{2} = 145$

Подставляем в начальное:

$\sqrt{5 - 1} + \sqrt{2 \times 5 - 1} = 5 \\ 2 + 3 = 5 \\ 5 = 5$

$\sqrt{145 - 1} + \sqrt{2 \times 145 - 1} = 5 \\ 12 + \sqrt{279} ≠5$

x=5