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Ma poate ajuta cineva, va rog?

Ma Poate Ajuta Cineva Va Rog class=

Răspuns :

[tex]\displaystyle a= \sqrt{12}-2 \sqrt{27}+5 \sqrt{3} + \sqrt{48}+2 \sqrt{75} \\ a=2 \sqrt{3}-2 \cdot 3 \sqrt{3} +5 \sqrt{3} +4 \sqrt{3} +2 \cdot 5 \sqrt{3} \\ a=2 \sqrt{3} -6 \sqrt{3} +5 \sqrt{3} +4 \sqrt{3} +10 \sqrt{3} \\ a=15 \sqrt{3 } [/tex]
[tex]\displaystyle b= \sqrt{(5-\sqrt{3})^2 } - \sqrt{(2 \sqrt{3}-3 \sqrt{2} )^2 } - \sqrt{(5-3 \sqrt{2} )^2} + \sqrt{7^2-1^2} \\ b=|5- \sqrt{3} |-|2 \sqrt{3}-3 \sqrt{2}|-|5-3 \sqrt{2}|+ \sqrt{49-1} \\ b=5- \sqrt{3} -(3 \sqrt{2}-2 \sqrt{3} )-(5-3 \sqrt{2} )+ \sqrt{48} \\ b=5- \sqrt{3} -3 \sqrt{2} +2 \sqrt{3} -5+3 \sqrt{2} +4 \sqrt{3} \\ b=5 \sqrt{3} \\ m_g= \sqrt{a \cdot b} = \sqrt{15 \sqrt{3} \cdot 5 \sqrt{3} }= \sqrt{15 \cdot 5 \cdot 3} = \sqrt{75 \cdot 3} = \sqrt{225}=15 \\ \boxed{m_g=15}[/tex]