Question

Помогите сделать задание 4.26

Answer 1

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
${( sqrt{6 - x} )}^{2} = {(2 sqrt{2} )}^{2} \ 6 - x = 4 times 2 \ 6 - x = 8 \ - x = 8 - 6 \ - x = 2 \ x = - 2$