Помогите пожалуйста!

1)
$\frac{1}{ \sqrt{26} -1} = \frac{(\sqrt{26}+1)}{ (\sqrt{26} -1)(\sqrt{26}+1)} =\frac{(\sqrt{26}+1)}{26-1} =\frac{(\sqrt{26}+1)}{25}$
2)
$\frac{35}{ \sqrt{37} + \sqrt{2} } = \frac{35( \sqrt{37} - \sqrt{2} )}{( \sqrt{37} + \sqrt{2})( \sqrt{37} - \sqrt{2} ) } = \frac{35( \sqrt{37} - \sqrt{2} )}{35 } =\sqrt{37} - \sqrt{2}$
3)
$\frac{16}{ \sqrt{47} - \sqrt{15} } = \frac{16( \sqrt{47} + \sqrt{15} ) }{( \sqrt{47} - \sqrt{15})( \sqrt{47} + \sqrt{15} ) } = \frac{16( \sqrt{47} + \sqrt{15} ) }{32 } = \frac{( \sqrt{47} + \sqrt{15} ) }{2 }$
4)
$\frac{x-4}{\sqrt{x+5}-3} = \frac{(x-4)(\sqrt{x+5}+3)}{(\sqrt{x+5}-3)(\sqrt{x+5}+3)} = \frac{(x-4)(\sqrt{x+5}+3)}{x+5-9} =\frac{(x-4)(\sqrt{x+5}+3)}{x-4} =$
$\sqrt{x+5}+3$

5)
$\frac{x^2+4x}{ \sqrt{x+8} -2} = \frac{(x^2+4x)(\sqrt{x+8} -2)}{ (\sqrt{x+8} -2)(\sqrt{x+8}+2)} =\frac{(x^2+4x)(\sqrt{x+8} -2)}{ x+8-4} =\frac{x(x+4)(\sqrt{x+8} -2)}{ x+4} =$
$x(\sqrt{x+8} -2)$