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July 2022 1 4 Report

Освободитесь от иррациональности в знаменатели дроби
[tex]\frac{4}{ \sqrt{3}+1 } } [/tex]
[tex]\frac{1}{ 1-\sqrt{2} } } [/tex]
[tex]\frac{4}{ \sqrt{10}- \sqrt{2} } } [/tex]
[tex]\frac{12}{ \sqrt{3}+\sqrt{6} } } [/tex]

Answers & Comments

mb17x $\frac{4}{ \sqrt{3} +1} = \frac{4( \sqrt{3} -1)}{( \sqrt{3}+1)( \sqrt{3} -1)} = \frac{4( \sqrt{3} -1)}{3-1} = \frac{4( \sqrt{3} -1)}{2} =2( \sqrt{3} -1)$
$\frac{1}{1- \sqrt{2} } = \frac{1+ \sqrt{2} }{(1- \sqrt{2})(1+ \sqrt{2} ) } = \frac{1+ \sqrt{2} }{1-2} =-1- \sqrt{2}$
$\frac{4}{ \sqrt{10}- \sqrt{2}}= \frac{4( \sqrt{10}+ \sqrt{2} )}{( \sqrt{10}+ \sqrt{2} )( \sqrt{10}- \sqrt{2} )} = \frac{4( \sqrt{10}+ \sqrt{2} )}{10-2} = \frac{4( \sqrt{10}+ \sqrt{2} )}{8}= \frac{ \sqrt{10}+ \sqrt{2} }{2}$
$\frac{12}{ \sqrt{3} + \sqrt{6}} = \frac{12(\sqrt{3}-\sqrt{6})}{(\sqrt{3} + \sqrt{6})(\sqrt{3} - \sqrt{6})} =\frac{12(\sqrt{3}-\sqrt{6})}{3-6} =\frac{12(\sqrt{3}-\sqrt{6})}{-3} =-4( \sqrt{3}- \sqrt{6})= \\ =4( \sqrt{6} - \sqrt{3} )$

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senia1257 https://znanija.com/task/22118485

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