Question

Ajutati.ma va rog urgent. Dau coroana

Answer 1

 

Nu se vede numerotarea exercitiilor.

Din acest motiv le voi numerota:  1)  si  2).

$\displaystyle\bf\\Ex.~1a)\\\\a,b,c~~dp~cu~~3,4,5\\\\\frac{a}{3}=\frac{b}{4}=\frac{c}{5}=k\\\\a=3k\\b=4k\\c=5k\\\\a+2b+3c=104\\\\3k+2\times4k+3\times5k=104\\\\3k+8k+15k=104\\\\26k=104\\\\k=\frac{104}{26}\\\\k=4\\\\a=3k=3\times4=12\\b=4k=4\times4=16\\c=5k=5\times4=20\\\\Ex.~1b)\\\\12^2+16^2=144+256=400=20^2\\\\\text{\bf Da, numerele a, b, c pot fi laturile unui triunghi dreptunghic.}$

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$\displaystyle\bf\\Ex.~2)\\\\a=\sqrt{12+6\sqrt{3}}-2\sqrt{\left(\sqrt{3}-2\right)^2}-3\sqrt{4-2\sqrt{3}}\\\\\\a=\sqrt{9+3+2\times3\times\sqrt{3}}-2\Big|\sqrt{3}-2\Big|-3\sqrt{3+1-2\times1\times\sqrt{3}}\\\\\\a=\sqrt{9+2\times3\times\sqrt{3}+3}-2\Big(2-\sqrt{3}\Big)-3\sqrt{3-2\times1\times\sqrt{3}+1}\\\\\\a=\sqrt{3^2+2\times3\times\sqrt{3}+(\sqrt{3})^2}-2\Big(2-\sqrt{3}\Big)-3\sqrt{(\sqrt{3})^2-2\times1\times\sqrt{3}+1}$

   

$\displaystyle\bf\\a=\sqrt{\Big(3+\sqrt{3}\Big)^2}-2\Big(2-\sqrt{3}\Big)-3\sqrt{\Big(\sqrt{3}-1\Big)^2}\\\\\\a=3+\sqrt{3}-2\Big(2-\sqrt{3}\Big)-3\Big(\sqrt{3}-1\Big)\\\\\\a=3+\sqrt{3}-4+2\sqrt{3}-3\sqrt{3}+3\\\\a=3-4+3+\sqrt{3}+2\sqrt{3}-3\sqrt{3}\\\\a=2+3\sqrt{3}-3\sqrt{3}\\\\a=2\in~N$