Answer:
1.
[tex] {( \sqrt{x - 1} })^{2} = {3}^{2} \\ x - 1 = 9 \\ x = 10[/tex]
2.
[tex] {( \sqrt{3x - 6}) }^{2} = {( \sqrt{2x - 12} )}^{2} \\ 3x - 6 = 2x - 12 \\ x = - 6[/tex]
3.
[tex] \frac{3 \sqrt{x} }{3} = \frac{6}{3} \\ {( \sqrt{x}) }^{2} = {2}^{2} \\ x = 4[/tex]
4.
[tex] \sqrt{x} = 12 - 8 \\ {( \sqrt{x} )}^{2} = {4}^{2} \\ x = 16[/tex]
5.
[tex] {( \sqrt{37 - 3x}) }^{2} = {(x - 3)}^{2} \\ 37 - 3x = {x}^{2} - 6x + 9 \\ {x}^{2} - 3x - 28 = 0 \\ x = 7[/tex]
6.
[tex] {( \sqrt{2x} )}^{2} = {6}^{2} \\ 2x = 36 \\ x = 18[/tex]
7.
[tex] {( \sqrt{ {x}^{2} - 5x - 10}) }^{2} = {x}^{2} \\ {x}^{2} - 5x - 10 = {x}^{2} \\ 5x = - 10 \\ x = - 2 [/tex]
8.
[tex] {( \sqrt{10 - {x}^{2} }) }^{2} = {(x + 2)}^{2} \\ 10 - {x}^{2} = {x}^{2} + 4x + 4 \\ 2 {x}^{2} + 4x - 6 = 0 \\ x = 1[/tex]
9.
[tex] {( \sqrt[3]{ {x}^{2} + 6x}) }^{3} = {3}^{3} \\ {x}^{2} + 6x = 27 \\ {x}^{2} + 6x - 27 = 0 \\ x = 3[/tex]
10.
[tex] {( \sqrt{2x - 1}) }^{2} = {(x - 8)}^{2} \\ 2x - 1 = {x}^{2} - 16x + 64 \\ {x}^{2} - 18x + 65 = 0 \\ x = 5 \: or \: 13[/tex]
