Question

Помогите решить 3 и 4 номер по алгебре

Answer 1

3a)$\frac{ \sqrt{10}+5 }{2+ \sqrt{10} } = \frac{ \sqrt{5*2}+( \sqrt{5}) ^{2} }{( \sqrt{2}) ^{2}+ \sqrt{2*5} } = \frac{ \sqrt{5}( \sqrt{2} + \sqrt{5} ) }{ \sqrt{2}( \sqrt{2} + \sqrt5)} }= \frac{ \sqrt{5} }{ \sqrt{2} } = \frac{ \sqrt{5} * \sqrt{2} }{ \sqrt{2}* \sqrt{2} } = \frac{ \sqrt{10} }{2}$
б) $\frac{x-3 \sqrt{x} }{2 \sqrt{x} -6} = \frac{( \sqrt{x} ) ^{2}-3 \sqrt{x} }{2( \sqrt{x} -3)}= \frac{ \sqrt{x} ( \sqrt{x} -3)}{2( \sqrt{x} -3)} = \frac{ \sqrt{x} }{2}$
4a)$\frac{7}{2 \sqrt{21} } = \frac{7* \sqrt{21} }{2 \sqrt{21} * \sqrt{21} }= \frac{7 \sqrt{21} }{42} = \frac{ \sqrt{21} }{6}$
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} )$