Помогите упростить и решить. Created by Raskolnikov1. algebra-ru

Помогите упростить и решить...

bayun1 $1) (\frac{ \sqrt{x} +1}{ \sqrt{x} -1} - \frac{ \sqrt{x} -1}{ \sqrt{x} +1} +4 \sqrt{x} ) ( \sqrt{x} - \frac{1}{ \sqrt{x} } ) = \\ \\ 
= \frac{( \sqrt{x} +1)( \sqrt{x} +1) - ( \sqrt{x}-1)( \sqrt{x} -1) +4 \sqrt{x} ( \sqrt{x} -1)( \sqrt{x} +1) }{( \sqrt{x} -1)( \sqrt{x} +1)} * \frac{ \sqrt{x} * \sqrt{x} -1}{ \sqrt{x} } = \\ \\ 
= \frac{( \sqrt{x} +1)^2 -( \sqrt{x} -1)^2 + 4 \sqrt{x} ( ( \sqrt{x} )^2 -1^2)}{( \sqrt{x} )^2-1^2} * \frac{x-1}{ \sqrt{x} } = \\ \\ 
$
$= \frac{x+2 \sqrt{x} +1 - (x-2 \sqrt{x} +1) + 4x \sqrt{x} -4 \sqrt{x} }{x-1} * \frac{x-1}{ \sqrt{x} } = \\ \\ 
= \frac{x+2 \sqrt{x} +1-x+2 \sqrt{x} -1+4x \sqrt{x} -4 \sqrt{x} }{x-1} * \frac{x-1}{ \sqrt{x} } = \\ \\ 
= \frac{4 x \sqrt{x} }{x-1} * \frac{x-1}{ \sqrt{x} } = 4x$

$2) (\frac{ \sqrt{x^2-4} - x}{ \sqrt{x^2-4} + x} - \frac{ \sqrt{x^2-4} +x}{ \sqrt{x^2-4} - x}) : \sqrt{ \frac{x^2-4}{x} } = \\ \\ 
= \frac{( \sqrt{x^2-4} - x)( \sqrt{x^2-4} - x) - ( \sqrt{x^2-4}+ x)( \sqrt{x^2-4} + x)}{( \sqrt{x^2-4}+ x)( \sqrt{x^2-4} - x)} : \sqrt{ \frac{x^2-4}{x} }= \\ \\ 
= \frac{( \sqrt{x^2-4} )^2 - 2x \sqrt{x^2-4 }+x^2-( \sqrt{x^2-4})^2-2x \sqrt{x^2-4} - x^2 }{( \sqrt{x^2-4})^2 -x^2 } : \sqrt{ \frac{x^2-4}{x} }= \\ \\ 
$
$= \frac{-4x \sqrt{x^2-4} }{x^2-4-x^2} :\sqrt{ \frac{1}{x} (x^2-4)}= \\ \\ 
= \frac{-4x \sqrt{x^2-4} }{-4} : \sqrt{ \frac{1}{x} } \sqrt{x^2-4}= \\ \\ 
= x \sqrt{x^2-4} : \sqrt{ \frac{1}{x} } \sqrt{x^2-4} = \\ \\ 
= \frac{x \sqrt{x^2-4} }{ \sqrt{ \frac{1}{x}}\sqrt{x^2-4} } = \frac{x}{ \sqrt{ \frac{1}{x} } } = \frac{x}{ \sqrt{x^{-1} } }$

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