Question

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Answer 1

Решение во вложении:

Answer 2

№7.

а)$\frac{22}{5+\sqrt{3} } =\frac{22(5-\sqrt{3}) }{(5+\sqrt{3} )(5-\sqrt{3}) } =\frac{22(5-\sqrt{3})}{25-3} =5-\sqrt{3}$

б) $\frac{12}{\sqrt{15} -\sqrt{5} } =\frac{12(\sqrt{15}+\sqrt{5} )}{(\sqrt{15} -\sqrt{5})(\sqrt{15} +\sqrt{5} ) }=\frac{12(\sqrt{15}+\sqrt{5} )}{15-5} =1,2(\sqrt{15}+\sqrt{5} )$

№8.

а) $\sqrt{m^7} =\sqrt{m^6*m}= m^3\sqrt{m}$

б) $\sqrt{7t^{18}} ; t<0\\\sqrt{7t^{18}}=-t^9\sqrt{7}$

№9.

$(\sqrt{5-2\sqrt{6} } -\sqrt{5+2\sqrt{6} } )^2=(\sqrt{2-2\sqrt{6}+3 }-\sqrt{2+2\sqrt{6} +3} )^2=(\sqrt{(\sqrt{2}-\sqrt{3})^2 }-\sqrt{(\sqrt{2}+\sqrt{3} )^2 } =((\sqrt{3} -\sqrt{2}) -(\sqrt{2}+ \sqrt{3} ))^2=(\sqrt{3} -\sqrt{2} -\sqrt{2} -\sqrt{3} )^2=(-2\sqrt{2} )^2=4*2=8$