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

а) [latex] \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} =[/latex] [latex]= \frac{(2-5) \sqrt{10} }{-6}= \frac{-3 \sqrt{10} }{-6} = \frac{ \sqrt{10} }{2} [/latex] б) [latex] \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)} =[/latex] [latex]= \frac{(2x-18) \sqrt{x} }{2(2x-18)} = \frac{ \sqrt{x} }{2} [/latex] 2. a) [latex] \frac{7}{2 \sqrt{21} } = \frac{7\sqrt{21} }{2 \sqrt{21}* \sqrt{21} }= \frac{7 \sqrt{21} }{2*21} = \frac{ \sqrt{21} }{6} [/latex] б) [latex] \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} =[/latex] [latex]=2( \sqrt{13}+ \sqrt{2}). [/latex]