Question

потрібно допомога.......​

Answer 1

Ответ:

#7.

a) $\frac{23}{5 - \sqrt{2} } = \frac{23}{\sqrt{25 - 2} } = \frac{23}{\sqrt{23} } = \frac{23^{2} }{23} = 23$

б) $\frac{15}{\sqrt{11} + \sqrt{5} }$ = $\frac{15 * (\sqrt{11} - \sqrt{5} ) }{(\sqrt{11} - \sqrt{5} ) * (\sqrt{11} + \sqrt{5} )}$   = $\frac{15 * (\sqrt{11} - \sqrt{5} ) }{11-5}$ = $\frac{15 * (\sqrt{11} - \sqrt{5} ) }{6}$ = $\frac{5 * (\sqrt{11} - \sqrt{5} ) }{2}$

#8.

a) не видно ступінь

б) $\sqrt{2b^{14}} = b^7\sqrt{2}$

#9.

$(\sqrt{5 + 2\sqrt{6} } + \sqrt{5 - 2\sqrt{6} })^2 =$ $(\sqrt{2 + 2\sqrt{2}*\sqrt{3} + 3} + \sqrt{2 - 2\sqrt{2}*\sqrt{3} + 3})^2$ = $(\sqrt{(\sqrt{2} + \sqrt{3} )^2 } + (\sqrt{(\sqrt{2} - \sqrt{3} )^2 })^2$ = $((\sqrt{2} + \sqrt{3} ) - (\sqrt{2} - \sqrt{3} }))^2$ = $(\sqrt{2} + \sqrt{3} -\sqrt{2} + \sqrt{3} })^2 = (2\sqrt{3} )^2 = 4 * 3 = 12$