Помогите решить. [tex] \frac{(5 \sqrt{3}+ \sqrt{50})(5- \sqrt{24})}{ \sqrt{75}-5 \sqrt{2} } [/tex]. Created by assema1998. algebra-ru

tex]...

julyap
$\frac{(5 \sqrt{3}+ \sqrt{50} )(5- \sqrt{24}) }{ \sqrt{75} -5 \sqrt{2} }=\frac{5\sqrt{3} -5 \sqrt{2} }{5\sqrt{3} -5 \sqrt{2} } =1$

$\\ \\ (5 \sqrt{3} + \sqrt{50} )(5- \sqrt{24} )= \\ =25 \sqrt{3} -5 \sqrt{72} +5 \sqrt{50} - \sqrt{1200} = \\ =25 \sqrt{3} -5 \sqrt{36*2} +5 \sqrt{25*5} - \\ - \sqrt{400*3} =25 \sqrt{3} -5*6 \sqrt{2} + \\ +5*5 \sqrt{2} -20 \sqrt{3} =25 \sqrt{3} -30 \sqrt{2} + \\ +25 \sqrt{2} -20 \sqrt{3} =25 \sqrt{3} -20 \sqrt{3} + \\ +25 \sqrt{2} -30 \sqrt{2} =5 \sqrt{3} -5 \sqrt{2} \\ \\ \sqrt{75} -5 \sqrt{2} = \sqrt{25*3} -5 \sqrt{2} = \\ =5 \sqrt{3} -5 \sqrt{2} \\ \\$

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julyap 1 строчка

kazimierz2015 $\frac{(5 \sqrt{3}+ \sqrt{50 })(5- \sqrt{24}) }{ \sqrt{75} -5 \sqrt{2} } = \\ 
= \frac{(5 \sqrt{3}+5 \sqrt{2})(5-2 \sqrt{6}) }{5 \sqrt{3}-5 \sqrt{2} } } =
 \frac{(5 \sqrt{3}+5 \sqrt{2})^2(5-2 \sqrt{6} ) }{25*3-25*2} = \\ 
= \frac{(75+50 \sqrt{6}+50)(5-2 \sqrt{6} ) }{75-50} 
= \frac{(125+50 \sqrt{6})(5-2 \sqrt{6} ) }{25}= \\ 
= \frac{625-250 \sqrt{6}+250 \sqrt{6}-600 }{25}=
 \frac{25}{25}=1$

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