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

Răspuns :

[tex] \sqrt{48}-2 \sqrt{3}*(2-5 \sqrt{12} )=4\sqrt{3}-2 \sqrt{3}*(2-5 *2\sqrt{3} )=[/tex]
=[tex]4\sqrt{3}-2 \sqrt{3}*(2-10\sqrt{3} )=4\sqrt{3}-4 \sqrt{3}+20*3=60[/tex]

[tex](3+ \sqrt{3} )*(2+ \sqrt{3} )=6+3 \sqrt{3} +2 \sqrt{3}+3=9+5 \sqrt{3} [/tex][tex] ( \sqrt{6}+ \sqrt{5} )^{2} - \sqrt{120} =(\sqrt{6})^{2}+2*\sqrt{6}*\sqrt{5}+(\sqrt{5})^{2}-2 \sqrt{30}= [/tex]
[tex]=6+2 \sqrt{30} +5-2 \sqrt{30}=11[/tex]

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