Question

це остане питання !!!!!!!!!! брат помги последний раз

Answer 1

Объяснение:

7.

$\displaystyle\\1)\ \frac{7}{\sqrt{14} }= \frac{7*\sqrt{14} }{\sqrt{14}*\sqrt{14} } =\frac{7*\sqrt{14} }{14} =\frac{\sqrt{14} }{2}.$

$\displaystyle\\2)\ \frac{2}{\sqrt{11}-3 }=\frac{2*(\sqrt{11}+3) }{(\sqrt{11}-3)(\sqrt{11}+3) } =\frac{2*(\sqrt{11}+3) }{11-9 }=\frac{2*(\sqrt{11}+3) }{2 }=\\\\\\=\sqrt{11}+3.$

8.

$\displaystyle\\\frac{\sqrt{45}+\sqrt{18} }{\sqrt{5}-\sqrt{2} }+ \frac{\sqrt{45}-\sqrt{18} }+{\sqrt{5}+\sqrt{2} }=\frac{\sqrt{9*5}+\sqrt{9*2} }{\sqrt{5}-\sqrt{2} } +\frac{\sqrt{9*5}-\sqrt{9*2} }{\sqrt{5}+\sqrt{2} } =\\\\\\=\frac{3\sqrt{5}+3\sqrt{2} }{\sqrt{5} -\sqrt{2} } +\frac{3\sqrt{5}-3\sqrt{2} }{\sqrt{5} +\sqrt{2} } =\frac{3*(\sqrt{5}+\sqrt{2} ) }{\sqrt{5}-\sqrt{2} }+\frac{3*(\sqrt{5}-\sqrt{2} ) }{\sqrt{5}+\sqrt{2} }=$

$\displaystyle\\=\frac{3*(\sqrt{5}+\sqrt{2})^2+3*(\sqrt{5}-\sqrt{2}) }{(\sqrt{5}-\sqrt{2})*(\sqrt{5}+\sqrt{2}) }=\frac{3*((\sqrt{5}+\sqrt{2})^2+(\sqrt{5}-\sqrt{2})^2) }{(\sqrt{5})^2-(\sqrt{2})^2 } =\\\\\\=\frac{3*((\sqrt{5}+\sqrt{2})^2+(\sqrt{5}-\sqrt{2})^2) }{5-2 } =\frac{3*((\sqrt{5}+\sqrt{2})^2+(\sqrt{5}-\sqrt{2})^2) }{3 } =\\\\\\=(\sqrt{5}+\sqrt{2})^2+(\sqrt{5}-\sqrt{2})^2=\\\\\\=(\sqrt{5})^2+2*\sqrt{5}*\sqrt{2}+(\sqrt{2})^2 +(\sqrt{5})^2-2*\sqrt{5}*\sqrt{2} +(\sqrt{2)^2} =$

$=5+2+5+2=14.$