Question

Помогите сделать задания

Answer 1

a) $\sqrt{x+5} +1= x$

$\sqrt{x+5} = x-1$

$(\sqrt{x+5})^{2} = (x-1)^{2}$

$\left \{ {{x+5=x^{2}-2x+1 } \atop {x+1>0 }} \right.$

$\left \{ {{x^{2}-3x-4=0 } \atop {x>-1 }} \right.$

x² - 3x - 4 = 0

D = (-3)² - 4 * ( -4) = 9 + 16 = 25 = 5²

$x_{1} = \frac{3+5}{2*1} = \frac{8}{2} = 4$

$x_{2} = \frac{3-5}{2*1} =- \frac{2}{2} = -1$ - не подходит под ОДЗ

Ответ: х = 4

б) $\sqrt{x+3} + \sqrt{3x-2} = 7$

$(\sqrt{x+3} + \sqrt{3x-2})^{2} = (7)^{2}$

$x+3 + 2*\sqrt{(x+3)(3x-2)} +3x-2 = 49$

$2*\sqrt{(x+3)(3x-2)} = 48 - 4x$ | :2

$\sqrt{3x^{2}+7x-6 } = 24 - 2x$

$(\sqrt{3x^{2}+7x-6 })^{2} = (24 - 2x)^{2}$

$\left \{ {{3x^{2}+7x-6 } = 576 - 96x + 4x^{2} } \atop {24-2x>0}} \right.$

$\left \{ {{x^{2}-103x+582=0 } \atop {x<12}} \right.$

x² - 103x +582 = 0

По т. Виета:

x₁ + x₂ = -b

x₁ * x₂ = c

x₁ = 97 - не подходит под ОДЗ

x₂ = 6

Ответ: х = 6

в) x² + 10x + 9 = 0

D = 10² - 4 * 9 = 100 - 36 = 64 = 8²

$x_{1} = \frac{-10+8}{2*1} = -\frac{2}{2} = -1$

$x_{2} = \frac{-10-8}{2*1} = -\frac{18}{2} = -9$

Ответ: х₁ = -9 , x₂ = -1