Question

помогите пожалуйста срочно
20 балов​

Answer 1

Ответ:

1.

А)

${x}^{2} = 25 \\ {x}^{2} - 25 = 0 \\ (x - 5)(x + 5) = 0 \\ 1)x - 5 = 0 \\ x = 5 \\ 2)x + 5 = 0 \\ x = - 5$

Б)

${y}^{2} = 19 \\ {y}^{2} - 19 = 0 \\ (y - \sqrt{19} )(y + \sqrt{19} ) = 0 \\ 1)y - \sqrt{19} = 0 \\ y = \sqrt{19} \\ 2)y + \sqrt{19} = 0 \\ y = - \sqrt{19}$

В)

${x}^{2} = - 16 \\ {x}^{2} + 16 = 0$

не имеет действительных корней

Г)

${x}^{2} = 0.49 \\ {x}^{2} - 0.49 = 0 \\ (x - 0.7)(x + 0.7) = 0 \\ 1)x - 0.7 = 0 \\ x = 0.7 \\ 2)x + 0.7 = 0 \\ x = - 0.7$

2.

А)

$( \sqrt{5} - \sqrt{2} )^{2} = { (\sqrt{5}) }^{2} - 2 \times \sqrt{5 \times 2} + {( \sqrt{2}) }^{2} = 5 - 2 \sqrt{10} + 2 = 7 - 2 \sqrt{10}$

Б)

$( \sqrt{10} + \sqrt{6} )( \sqrt{10} - \sqrt{6} ) = {( \sqrt{10} )}^{2} - { (\sqrt{6}) }^{2} = 10 - 6 = 4$

3.

А)

$\frac{15}{ \sqrt{5} } = \frac{15 \times \sqrt{5} }{ \sqrt{5} \times \sqrt{5} } = \frac{15 \sqrt{5} }{5} = 3 \sqrt{5}$

Б)

$\frac{2}{3 \sqrt{7} } = \frac{2 \times \sqrt{7} }{3 \sqrt{7} \times \sqrt{7} } = \frac{2 \sqrt{7} }{3 \times 7} = \frac{2 \sqrt{7} }{21}$

В)

$\frac{5}{ \sqrt{13} - \sqrt{3} } = \frac{5 \times ( \sqrt{13} + \sqrt{3} ) }{( \sqrt{13} - \sqrt{3} )( \sqrt{13} + \sqrt{3}) } = \frac{5 \times ( \sqrt{13} + \sqrt{3} ) }{13 - 3} = \frac{5 \times ( \sqrt{13} + \sqrt{3} ) }{10} = \frac{ \sqrt{13} + \sqrt{3} }{2}$

Г)

$\frac{4}{ \sqrt{11} + 3} = \frac{4 \times ( \sqrt{11} - 3)}{ (\sqrt{11} + 3)( \sqrt{11} - 3) } = \frac{4 \times ( \sqrt{11} - 3)}{11 - 9} = \frac{4 \times ( \sqrt{11} - 3) }{2} = 2 \sqrt{11} - 6$

4.

$10 \sqrt{3} - 4 \sqrt{48} - \sqrt{75} = 10 \sqrt{3} - 4 \sqrt{16 \times 3} - \sqrt{25 \times 3} = 10 \sqrt{3} - 4 \times 4 \sqrt{3} - 5 \sqrt{3} = 10 \sqrt{3} - 16 \sqrt{3} - 5 \sqrt{3} = - 11 \sqrt{3}$