Question

ajutalima varog frumos
√48-2√3(2-5√12)
(3+√3)(2+√3)
(√6+√5)²-√120

Answer 1

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

$(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}=$
$=6+2 \sqrt{30} +5-2 \sqrt{30}=11$