Question

Помогите пожалуйста ​

Answer 1

Ответ:

1)

$\sqrt{ {x}^{2} + 5x - 6 } = x + 2$

$\sqrt{ {x}^{2} + 5x - 6 } ^{2} = (x + 2)^{2}$

${x}^{2} + 5x - 6 = {x}^{2} + 4x + 4$

$5x - 6 = 4x + 4$

$5x - 4x = 4 + 6$

$x = 10$

$\sqrt{ {10}^{2} + 5 \times 10 - 6 } = 10 + 2$

$\sqrt{144} = 12$

$12 = 12$

$x = 10$

2)

$2 \sqrt{x + 5} = x + 2$

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

$4(x + 5) = {x}^{2} + 4x + 4$

$4x + 20 = {x}^{2} + 4x + 4$

$20 = {x}^{2} + 4$

$- {x}^{2} = 4 - 20$

$- {x}^{2} = - 16$

${x}^{2} = 16$

$x = ± \sqrt{16}$

$x = ±4$

$x = - 4 \\ x = 4$

$2 \sqrt{ - 4 + 5} = - 4 + 2 \\ 2 \sqrt{4 +5 } = 4 + 2$

$2 \sqrt{ - 4 + 5} = - 4 + 2 \\ 2 \sqrt{1} = - 4 + 2 \\ 2 \sqrt{1} = - 2 \\ 2 \times 1 = - 2 \\ 2 = - 2$

$2 \sqrt{4 + 5} = 4 + 2 \\ 2 \sqrt{9} = 4 + 2 \\ 2 \sqrt{9} = 6 \\ 2 \times 3 = 6 \\ 6 = 6$

$2 = - 2 \\ 6 = 6$

$x≠ - 4 \\ x = 4$

$x = 4$

3)

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

$\sqrt{2 {x}^{2} + 7} = x + 2$

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

$2 {x}^{2} + 7 = {x}^{2} + 4x + 4$

$2x ^{2} + 7 - {x}^{2} - 4x - 4 = 0$

${x}^{2} + 3 - 4x = 0$

${x}^{2} - 4x + 3 = 0$

${x}^{2} - x - 3x + 3 = 0$

$x \times (x - 1) - 3(x - 1) = 0$

$(x - 1) \times (x - 3) = 0$

$x - 1 = 0 \\ x - 3 = 0$

$x - 1 = 0 \\ x - 1 + 1 = 0 + 1 \\ x = 1$

$x - 3 = 0 \\ x - 3 + 3 = 0 + 3 \\ x = 3$

$x = 1 \\ x = 3$

$\sqrt{2 \times 1 ^{2} + 7 } - 2 = 1 \\ \sqrt{2 \times {3}^{2} + 7 } - 2 = 3$

$\sqrt{2 \times {1}^{2} + 7} - 2 = 1 \\ \sqrt{2 \times 1 + 7} - 2 = 1 \\ \sqrt{2 + 7} - 2 = 1 \\ \sqrt{9} - 2 = 1 \\ 3 - 2 = 1 \\ 1 = 1$

$\sqrt{2 \times {3}^{2} + 7} - 2 = 3 \\ \sqrt{2 \times 9 + 7} - 2 = 3 \\ \sqrt{18 + 7} - 2 = 3 \\ \sqrt{25} - 2 = 3 \\ 5 - 2 = 3 \\ 3 = 3$

$1 = 1 \\ 3 = 3$

$x = 1 \\ x = 3$

$x _{1} = 1. x_{2} = 3$

4)

$\sqrt[5]{ {x}^{5} - 5 {x}^{2} + 5} = x$

$\sqrt[5]{ {x}^{5} - 5 {x}^{2} + 5} ^{5} = {x}^{5}$

${x}^{5} - 5 {x}^{2} + 5 = {x}^{5}$

$- 5 {x}^{2} + 5 = 0$

$- 5 {x}^{2} = - 5$

|:-5

${x}^{2} = 1$

$x = ± \sqrt{1}$

$x = ±1$

$x = - 1 \\ x = 1$

$x_{1} = - 1. x_{2} = 1$