Question

Va rog exercitiile din imagine)) Mersi..)

Answer 1

rezolvarea la primele 2 le gasesti in atasament

Answer 2

$1. a) 4 \sqrt{3} -3 \sqrt{2} -6 \sqrt{3} +8 \sqrt{2} =5 \sqrt{2} -2 \sqrt{3}$

$b) 9 \sqrt{2} -5 \sqrt{3}-4 \sqrt{2}+7 \sqrt{3}=5 \sqrt{2} +2 \sqrt{3}$

$a*b=(5 \sqrt{2}-2\sqrt{3})*(5 \sqrt{2}+2 \sqrt{3})= (5 \sqrt{2} )^{2} - (2 \sqrt{3} )^{2} =50-12=38$

$2.a) 2 \sqrt{5} -7 \sqrt{5}+16 \sqrt{5}+5 \sqrt{5} +2 \sqrt{5}=14 \sqrt{5}$

$b) 12 \sqrt{2} - 25 \sqrt{2} -11 \sqrt{2}+8 \sqrt{2} =-16 \sqrt{2}$

$c) 12 \sqrt{6} +2 \sqrt{3}*(4 \sqrt{3} -6 \sqrt{2} )=12 \sqrt{6} +24-12 \sqrt{6} =24$

$3. a) \frac{ \sqrt{24} }{7} *( \sqrt{ \frac{1}{16} } + \frac{2 \sqrt{3} }{ \sqrt{2} } )= \frac{ 2\sqrt{6} }{7} *( \frac{1}{4} + \frac{ 2\sqrt{3} }{ \sqrt{2} } )= \frac{2 \sqrt{6} }{7} * \frac{ \sqrt{2}+4 \sqrt{6} }{4} =$
$= \frac{2 \sqrt{6} }{7} * \frac{ \sqrt{2}(1+4 \sqrt{3}) }{4} = \frac{4 \sqrt{3}(1+4 \sqrt{3} ) }{28} = \frac{12+4 \sqrt{3} }{7}$

$b) ( \frac{2}{3 \sqrt{2} } + \frac{5}{2 \sqrt{2} } )-( \frac{3 \sqrt{2} }{2} - \frac{1}{3 \sqrt{2} } )= \frac{4 \sqrt{2}+15 \sqrt{2} }{12} - \frac{9 \sqrt{2}- \sqrt{2} }{6} = \frac{19 \sqrt{2} }{12} - \frac{8 \sqrt{2} }{6}$
$= \frac{19 \sqrt{2}-16 \sqrt{2} }{12} = \frac{3 \sqrt{2} }{12} = \frac{ \sqrt{2} }{4}$

$c) 12 \sqrt{6} +2 \sqrt{3} *(4 \sqrt{3} -6 \sqrt{2} )=12 \sqrt{6}+24-12 \sqrt{6} =24$