Question

Плиз хелп дам 40б?????​

Answer 1

Ответ:

a) 5

б) -34

Объяснение:

a)

$\displaystyle \frac{\sqrt{7} -\sqrt{3} }{\sqrt{7} +\sqrt{3} } ^{\setminus \sqrt{7}-\sqrt{3} } +\frac{\sqrt{7} +\sqrt{3} }{\sqrt{7} -\sqrt{3} } ^{\setminus \sqrt{7}+\sqrt{3} }=\frac{(\sqrt{7} -\sqrt{3})^2 +(\sqrt{7}+\sqrt{3} )^2)}{(\sqrt{7} )^2-(\sqrt{3}^2 )} =$

$\displaystyle \frac{(\sqrt{7})^2-2\sqrt{7} *\sqrt{3} +(\sqrt{3} ) ^2 + (\sqrt{7})^2+2\sqrt{7} *\sqrt{3} +(\sqrt{3} ) ^2 }{7-3} =\frac{7+3+7+3}{4} =5$

б)

$\displaystyle \frac{\sqrt{8}-3 }{\sqrt{8}+3 } +\frac{2\sqrt{2}+3 }{2\sqrt{2} -3} =\frac{(\sqrt{8}-3) ( 2\sqrt{2} -3)+ ( 2\sqrt{2} +3)(\sqrt{8} +3)}{(\sqrt{8} +3)(2\sqrt{2}-3) } =$

$\displaystyle =\frac{2\sqrt{8*2} -3\sqrt{8} -6\sqrt{2} +9+2\sqrt{2*8} +6\sqrt{2} +3\sqrt{8} +9}{(\sqrt{4*2} +3)(\sqrt{4*2} -3) } =\frac{4\sqrt{16} +18}{(\sqrt{8} )^2-3^2} =$

$\displaystyle =\frac{4*4+18}{8-9} =-34$