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