√13-√2...

June 2022 1 12 Report

Сократите дробь:
а) √10+5/2+√10
б) x-3√x/2√x-6
Освободите дробь от знака корня в знаменателе:
а) 7/2√21
б) 22/√13-√2

Answers & Comments

mayli

Verified answer

А) $\frac{ \sqrt{10}+5 }{2+ \sqrt{10} } = \frac{( \sqrt{10}+5)(2- \sqrt{10} ) }{(2+ \sqrt{10})(2- \sqrt{10}) } = \frac{2 \sqrt{10}-10+10-5 \sqrt{10} }{4-10} = \frac{2 \sqrt{10}-5 \sqrt{10} }{-6} =$

$= \frac{(2-5) \sqrt{10} }{-6}= \frac{-3 \sqrt{10} }{-6} = \frac{ \sqrt{10} }{2}$

б) $\frac{x-3 \sqrt{x} }{2 \sqrt{x} -6} = \frac{(x-3 \sqrt{x})(2 \sqrt{x} +6)}{(2 \sqrt{x} -6)(2 \sqrt{x} +6)} = \frac{2x \sqrt{x} +6x-6x-18 \sqrt{x} }{4x-36} = \frac{2x \sqrt{x} -18 \sqrt{x} }{2(2x-18)} =$

$= \frac{(2x-18) \sqrt{x} }{2(2x-18)} = \frac{ \sqrt{x} }{2}$

2.
a) $\frac{7}{2 \sqrt{21} } = \frac{7\sqrt{21} }{2 \sqrt{21}* \sqrt{21} }= \frac{7 \sqrt{21} }{2*21} = \frac{ \sqrt{21} }{6}$

б) $\frac{22}{ \sqrt{13}- \sqrt{2} } = \frac{22( \sqrt{13}+ \sqrt{2}) }{( \sqrt{13}- \sqrt{2}) (\sqrt{13}+ \sqrt{2})} = \frac{22( \sqrt{13}+ \sqrt{2})}{13-2} = \frac{22( \sqrt{13}+ \sqrt{2})}{11} =$

$=2( \sqrt{13}+ \sqrt{2}).$

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