Question

Помогите пожалуйста с 2 заданиями по Алгебре прошу

Answer 1

$3.1)10 \sqrt{3} - 4 \sqrt{48} - \sqrt{75} = \\ = 10 \sqrt{ 3} - 16 \sqrt{3} - 5 \sqrt{3} = \\ = - 11 \sqrt{3}$

$3.2)(5 \sqrt{2} - \sqrt{18} ) \times \sqrt{2} = \\ = (5 \sqrt{2} - 3 \sqrt{2} ) \times \sqrt{2} = \\ = 10 - 6 = 4$

$3.3) {(3 - \sqrt{2} )}^{2} = \\ = 9 - 6 \sqrt{2} + 2 = \\ =11 - 6 \sqrt{2}$

$3.4)( \sqrt{5} - \sqrt{18} )( \sqrt{18} + \sqrt{5} ) + \sqrt{225} = \\ = { \sqrt{5} }^{2} - { \sqrt{18} }^{2} + 15 = \\ = 5 - 18 + 15 = 2$

$4.1) \frac{6 + \sqrt{6} }{ \sqrt{6} } = \frac{ \sqrt{6}( \sqrt{6} + 1) }{ \sqrt{6} } = \sqrt{6 + 1}$

$\frac{9 - a}{3 + \sqrt{a} } = \frac{(3 - \sqrt{a})(3 + \sqrt{a}) }{3 + \sqrt{a} } = 3 - \sqrt{a}$