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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




Răspuns :

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

[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]