Помогите сделать задание 4.26

1.
$(x + 1) \sqrt{3} = x + 3 \\ x \sqrt{3} + \sqrt{3} = x + 3 \\ x \sqrt{3} - x = 3 - \sqrt{3} \\ x( \sqrt{3 } - 1) = 3 - \sqrt{3} \\ x = \frac{3 - \sqrt{3} }{ \sqrt{3} - 1 } \\ x = \frac{(3 - \sqrt{3})( \sqrt{3} + 1) }{ (\sqrt{3} - 1 )( \sqrt{3} + 1)} \\ x = \frac{3 \sqrt{3} + 3 - 3 - \sqrt{3} }{ {( \sqrt{3} )}^{2} - {1}^{2} } \\ x = \frac{2 \sqrt{3} }{3 - 1} = \frac{2 \sqrt{3} }{2} = \sqrt{3}$
2.
$(x - 1) \sqrt{2} = 2x - 1 \\ x \sqrt{2} - \sqrt{2} = 2x - 1 \\ x \sqrt{2} - 2x = \sqrt{2} - 1 \\ x( \sqrt{2} - 2) = \sqrt{2} - 1 \\ x = \frac{ \sqrt{2} - 1}{ \sqrt{2} - 2 } \\ x = \frac{ (\sqrt{2} - 1)( \sqrt{2} + 2) }{ (\sqrt{2} - 2)( \sqrt{2} + 2) } \\ x = \frac{2 + 2 \sqrt{2} - \sqrt{2} - 2 }{ {( \sqrt{2} )}^{2} - 2 {}^{2} } \\ x = \frac{ \sqrt{2} }{2 - 4} = \frac{ \sqrt{2} }{ - 2} = - \frac{ \sqrt{2} }{2}$
3.
$\sqrt{7x - 2} = 2$
одз: 7х-2≥0
7х≥2
х≥2/7

${ (\sqrt{7x - 2} )}^{2} = {2}^{2} \\ 7x - 2 = 4 \\ 7x = 4 + 2 \\ 7x = 6 \\ x = \frac{6}{7}$
4.
$\sqrt{6 - x} = 2 \sqrt{2}$
одз: 6-х≥0
х≤6
${( \sqrt{6 - x} )}^{2} = {(2 \sqrt{2} )}^{2} \\ 6 - x = 4 \times 2 \\ 6 - x = 8 \\ - x = 8 - 6 \\ - x = 2 \\ x = - 2$