Question

Освободитесь от иррациональности в знаменатели дроби
$frac{4}{ sqrt{3}+1 } }$
$frac{1}{ 1-sqrt{2} } }$
$frac{4}{ sqrt{10}- sqrt{2} } }$
$frac{12}{ sqrt{3}+sqrt{6} } }$

Answer 1

$frac{4}{ sqrt{3} +1} = frac{4( sqrt{3} -1)}{( sqrt{3}+1)( sqrt{3} -1)} = frac{4( sqrt{3} -1)}{3-1} = frac{4( sqrt{3} -1)}{2} =2( sqrt{3} -1)$
$frac{1}{1- sqrt{2} } = frac{1+ sqrt{2} }{(1- sqrt{2})(1+ sqrt{2} ) } = frac{1+ sqrt{2} }{1-2} =-1- sqrt{2}$
$frac{4}{ sqrt{10}- sqrt{2}}= frac{4( sqrt{10}+ sqrt{2} )}{( sqrt{10}+ sqrt{2} )( sqrt{10}- sqrt{2} )} = frac{4( sqrt{10}+ sqrt{2} )}{10-2} = frac{4( sqrt{10}+ sqrt{2} )}{8}= frac{ sqrt{10}+ sqrt{2} }{2}$
$frac{12}{ sqrt{3} + sqrt{6}} = frac{12(sqrt{3}-sqrt{6})}{(sqrt{3} + sqrt{6})(sqrt{3} - sqrt{6})} =frac{12(sqrt{3}-sqrt{6})}{3-6} =frac{12(sqrt{3}-sqrt{6})}{-3} =-4( sqrt{3}- sqrt{6})= \ =4( sqrt{6} - sqrt{3} )$

Answer 2

https://znanija.com/task/22118485