Question

Решите уравнение 1) и 5)

Answer 1

1.
$\sqrt{ {x}^{2} + 3x - 3} = 2x - 3 \\ {x}^{2} + 3x - 3 = 4 {x}^{2} - 12x + 9 \\ {x}^{2} + 3x - 3 - 4 {x}^{2} + 12x - 9 = 0 \\ - 3 {x}^{2} + 15x - 12 = 0 \\ {x}^{2} - 5x + 4 = 0 \\ d = b {}^{2} - 4ac = ( - 5) {}^{2} - 4 \times 1 \times 4 = 25 - 16 = 9 \\ x12 = \frac{ - b + - \sqrt{d} }{2a} = \frac{5 + - 3}{2} = \\ \frac{5 + 3}{2} = \frac{8}{2} = 4 \\ \frac{5 - 3}{2} = \frac{2}{2} = 1$
Проверка:
$\sqrt{ {4}^{2} + 3 \times 4 - 3 } = 2 \times 4 - 3 \\ \sqrt{ {1}^{2} + 3 \times 1 - 3} = 2 \times 1 - 3 \\ 5 = 5 \\ 1 = - 1$
Второе не верно, поэтому х=4.


2.
$\sqrt{4 - x} + \sqrt{5 + x} = 3 \\ 4 - x + 2 \sqrt{(4 - x) \times (5 + x)} + 5 + x = 9 \\ 4 + 2 \sqrt{20 + 4x - 5x - x {}^{2} } + 5 = 9 \\ 9 + 2 \sqrt{20 - x - x {}^{2} } = 9 \\ 2 \sqrt{20 - x - {x}^{2} } = 0 \\ \sqrt{20 - x - {x}^{2} } = 0 \\ 20 - x - x {}^{2} = 0 \\ - {x}^{2} - x + 20 = 0 \\ {x}^{2} + x - 20 = 0 \\ d = b {}^{2} - 4ac = {1}^{2} - 4 \times 1 \times ( - 20) = 1 - ( - 80) = 1 + 80 = 81 \\ x12 = \frac{ - b + - \sqrt{d} }{2a} = \frac{ - 1 + - 9}{2} = \\ \frac{ - 1 + 9}{2} = \frac{8}{2} = 4 \\ \frac{ - 1 - 9}{2} = - \frac{ 10}{2} = - 5$