Question

Ребят, пожалуйста, помогите ❤️

Answer 1

Ответ:

$10 \sqrt{3} - \sqrt{48} - \sqrt{75} = 10 \sqrt{3} - 4 \sqrt{3} - 5 \sqrt{3} = \sqrt{3}$

$( \sqrt{2} - \sqrt{18} ) \times \sqrt{2} = ( \sqrt{2} - 3 \sqrt{2} ) \times \sqrt{2} = - 2 \sqrt{2} \times \sqrt{2} = - 4$

$(3 - \sqrt{5} ) {}^{2} + 6 \sqrt{5} = (9 - 5) + 6 \sqrt{2} = 10 \sqrt{2}$

$\frac{a {}^{2} - 2 }{a + \sqrt{2} } = \frac{(a + \sqrt{2}) \times (a - \sqrt{2} )}{a + \sqrt{2} } = a - \sqrt{2}$

$2 \sqrt{2} + \sqrt{50} - \sqrt{32} = 2 \sqrt{2} + 5 \sqrt{2} - 4 \sqrt{2} = 3 \sqrt{2}$

$\sqrt{5} \times ( \sqrt{5} - \sqrt{20} ) = \sqrt{5} \times ( \sqrt{5} - 2 \sqrt{5} ) = \sqrt{5} \times ( - \sqrt{5} ) = - 5$

$( \sqrt{3 } + \sqrt{2} ) {}^{2} - 2 \sqrt{6} = (3 + 2 \sqrt{6} + 2) - 2 \sqrt{6} = 5$

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