tex]...

July 2022 1 10 Report

вычислить:
[tex] \sqrt{15+2 \sqrt{26} } * \sqrt{4+ \sqrt{1+2 \sqrt{26} } }*\sqrt{4- \sqrt{1+2 \sqrt{26} } }[/tex]
[tex] \sqrt{ \frac{11+ \sqrt{21} }{11- \sqrt{21} } } + \sqrt{ \frac{11- \sqrt{21} }{11+ \sqrt{21} } } [/tex]
разложить на множители
m+[tex] \sqrt{m}-6 [/tex]
2y+[tex] \sqrt{y}-3[/tex]

Answers & Comments

eugeke

Verified answer

$\sqrt{(15+2\sqrt{26}}\cdot \sqrt{4+\sqrt{1+2\sqrt{26}}}\cdot \sqrt{4-\sqrt{1+2\sqrt{26}}}} =\\
 \sqrt{(15+2\sqrt{26})(4-\sqrt{1+2\sqrt{26})}(4-\sqrt{1+2\sqrt{26})}} =\\
 \sqrt{(15+2\sqrt{26})( 16-( \sqrt{1+2 \sqrt{26} } )^2)}= \sqrt{(15+2\sqrt{26})(15-2 \sqrt{26} )}} =\\
 \sqrt{15^2-(2\sqrt{26})^2}= \sqrt{225-104}= \sqrt{121}=11$

$\sqrt{ \frac{11+ \sqrt{21} }{11- \sqrt{21}} } + \sqrt{ \frac{11- \sqrt{21} }{11+\sqrt{21}}}= \frac{ (\sqrt{11+ \sqrt{21}})^2+(\sqrt{11- \sqrt{21}})^2 }{ \sqrt{(11- \sqrt{21})(11+ \sqrt{21})^ }}}=\\\\
 \frac{11+ \sqrt{21}+11-\sqrt{21} }{ \sqrt{11^2-( \sqrt{21} )^2} }= \frac{22}{ \sqrt{100}}= \frac{22}{10}=2,2$

$m+ \sqrt{m}-6= \sqrt{m} \sqrt{m} + \sqrt{m}-2\cdot3=( \sqrt{m}+3)( \sqrt{m}-2)$

$2y+ \sqrt{y}-3=2 \sqrt{y} \sqrt{y}+ \sqrt{y}-3\cdot1= (2 \sqrt{y}+3)( \sqrt{y}-1)$

3 votes Thanks 2

taniachaika61 спасибо большое

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