Question

ПОЖАЛУЙСТА ПОМОГИТЕ РЕШИТЬ СРОЧНО!!!!!​

Answer 1

Ответ:

$\sqrt{ {x}^{2} + 3x - 2 } = \sqrt{ - 2x - 1} \\ (\sqrt{ {x}^{2} + 3x - 2 })^{2} = (\sqrt{ - 2x - 1})^{2} \\ {x}^{2} + 3x - 2 = - 2x - 1 \\ {x}^{2} + 3x - 2 + 2x + 1 = 0 \\ {x}^{2} + 5x - 1 = 0 \\ d = {b}^{2} - 4ac = {5}^{2} - 4 \times 1 \times ( - 1) = 25 + 4 = 29 \\ x1 = \frac{ - b - \sqrt{d} }{2a} = \frac{ - 5 - \sqrt{29} }{2} \\ x2 = \frac{ - b + \sqrt{d} }{2a} = \frac{ - 5 + \sqrt{29} }{2}$

$\sqrt{ {x}^{2} + 8x - 7} = \sqrt{ - x - 1} \\ (\sqrt{ {x}^{2} + 8x - 7})^{2} = (\sqrt{ - x - 1})^{2} \\ {x}^{2} + 8x - 7 = - x - 1 \\ {x}^{2} + 8x - 7 + x + 1 = 0 \\ {x}^{2} + 9x - 6 = 0 \\ d = {b}^{2} - 4ac = {9}^{2} - 4 \times 1 \times ( - 6) = 81 + 24 = 105 \\ x1 = \frac{ - b - \sqrt{d} }{2a} = \frac{ - 9 - \sqrt{105} }{2} \\ x2 = \frac{ - b + \sqrt{d} }{2a} = \frac{ - 9 + \sqrt{105} }{2}$

$\sqrt{ {x}^{2} - 6x + 2 } = x + 5 \\ (\sqrt{ {x}^{2} - 6x + 2 })^{2} =( x + 5)^{2} \\ {x}^{2} - 6x + 2 = {x}^{2} + 10x + 25 \\ {x}^{2} - 6x + 2 - {x}^{2} - 10x - 25 = 0 \\ - 16x - 23 = 0 \\ - 16x = 23 \\ x = - 23 \div 16 \\ x = - 1 \frac{7}{16}$

$\sqrt{ - 2 {x}^{2} - 3x - 2} = \sqrt{x + 1} \\ (\sqrt{ - 2 {x}^{2} - 3x - 2} )^{2} = (\sqrt{x + 1}) ^{2} \\ - 2 {x}^{2} - 3x - 2 = x + 1 \\ - 2 {x}^{2} - 3x - 2 - x - 1 = 0 \\ - 2 {x}^{2} - 4x - 3 = 0 \\ 2 {x}^{2} + 4x + 3 = 0 \\ d = {b}^{2} - 4ac = {4}^{2} - 4 \times 2 \times 3 = 16 - 24 = - 8$

Нет решений, так как дискриминант < 0

$\sqrt{ - {x}^{2} + 3x + 5 } = \sqrt{x + 10} \\ (\sqrt{ - {x}^{2} + 3x + 5 } )^{2} = (\sqrt{x + 10}) ^{2} \\ - {x}^{2} + 3x + 5 = x + 10 \\ - {x}^{2} + 3x + 5 - x - 10 = 0 \\ - {x}^{2} + 2x - 5 = 0 \\ {x}^{2} - 2x + 5 = 0 \\ d = {b}^{2} - 4ac = {( - 2)}^{2} - 4 \times 1 \times 5 = 4 - 20 = - 16$

Нет решений, так как дискриминант < 0