Question

am nevoie de rezolvare 
√3(√12 - 2√27) ; 
√48-2√3(2-5√12) ; 
(3+√3)(2+√3) ; 
(√6+√5)∧2-√120 ; 
cine stie , clasa 9 ex.10 pag.18

Answer 1

[tex] \sqrt{3}*(\sqrt{12}-2 \sqrt{27} )= \sqrt{3}*(2\sqrt{3}-2 *3\sqrt{3} )=\sqrt{3}*(2\sqrt{3}-6\sqrt{3} )= [/tex]
$=\sqrt{3}*(-4\sqrt{3})=-4*3=-12$

$\sqrt{48}-2 \sqrt{3} *(2-5 \sqrt{12} )=4 \sqrt{3} -2 \sqrt{3}*(2-5*2 \sqrt{3} ) =$
$=4 \sqrt{3} -2 \sqrt{3}*(2-10 \sqrt{3} ) =4 \sqrt{3}-4 \sqrt{3}+20*3=60$

$(3+ \sqrt{3} )*(2+ \sqrt{3} )=6+3 \sqrt{3}+2 \sqrt{3} +3 =9+5 \sqrt{3}$

$( \sqrt{6} +\sqrt{5} )^{2} - \sqrt{120}=( \sqrt{6} ) ^{2}+2* \sqrt{6} * \sqrt{5} + ( \sqrt{5} )  ^{2}-2 \sqrt{30}=$
$=6+2 \sqrt{30} +5-2 \sqrt{30}=11$