Question

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Answer 1

  $\displaystyle\bf\\2)\\a=4\sqrt{z}\\b=2\sqrt{z}\\\\m_a=\frac{a+b}{2}=\frac{4\sqrt{z}+2\sqrt{z}}{2}=\frac{6\sqrt{z}}{2}=3\sqrt{z} \\\\m_g=\sqrt{a\times b}=\sqrt{4\sqrt{z}\times2\sqrt{z}}=\sqrt{8z}=2\sqrt{2z}$

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$\displaystyle\bf\\3)\\a)~~Rationalizam~numitorul:\\\\\frac{^{\sqrt{3})}5}{4\sqrt{3}}=\frac{5\sqrt{3}}{4\sqrt{3}\times\sqrt{3}}=\frac{5\sqrt{3}}{4\times3}=\frac{5\sqrt{3}}{12}\\\\\\b)Calculam:\\\\2\sqrt{8}-5\sqrt{32}+4\sqrt{2}=\\\\=2\sqrt{4\times2}-5\sqrt{16\times2}+4\sqrt{2}=\\\\=2\times2\sqrt{2}-5\times4\sqrt{2}+4\sqrt{2}=\\\\=4\sqrt{2}-20\sqrt{2}+4\sqrt{2}=-12\sqrt{2}$

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$\displaystyle\bf\\4)~Calculam:\\\\~~~|3\sqrt{7}-8| +~|5-2\sqrt{7}~|\\\\In~fiecare~modul~comparam~termenii~prin~ridicare~la~patrat:\\\\\Big(3\sqrt{7}\Big)^2=9\times7=63\\\\8^2=64\\\\\implies~8>3\sqrt{7}\\\\\implies~|3\sqrt{7}-8|=8-3\sqrt{7}\\\\5^2=25\\\\\Big(2\sqrt{7}\Big)^2=4\times7=28\\\\\implies~2\sqrt{7}>5\\\\\implies~|5-2\sqrt{7}~|=2\sqrt{7}-5\\\\|3\sqrt{7}-8| +~|5-2\sqrt{7}~|=8-3\sqrt{7}+2\sqrt{7}-5=3-\sqrt{7}$

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$\displaystyle\bf\\5) Stabilim~daca:\\\\\left[\frac{5}{\sqrt{12}}-\sqrt{27}^{~-\b1}+\left(-\frac{1}{3\sqrt{3}}\right)\right]\times\sqrt{108}~\in N\\\\\\Rezolvare:\\\\\left[\frac{5}{\sqrt{12}}-\sqrt{27}^{~-\b1}+\left(-\frac{1}{3\sqrt{3}}\right)\right]\times\sqrt{108}=\\\\\\=\left[\frac{5}{\sqrt{12}}-\frac{1}{\sqrt{27}}-\frac{1}{3\sqrt{3}}\right]\times\sqrt{108}=\\\\\\=\left[\frac{5\times3}{3\times2\sqrt{3}}-\frac{2}{2\times3\sqrt{3}}-\frac{2}{2\times3\sqrt{3}}\right]\times6\sqrt{3}=$

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$\displaystyle\bf\\=\left[\frac{15}{6\sqrt{3}}-\frac{2}{6\sqrt{3}}-\frac{2}{6\sqrt{3}}\right]\times6\sqrt{3}=\\\\\\=\frac{15-2-2}{6\sqrt{3}}\times6\sqrt{3}=\\\\\\=\frac{11\times6\sqrt{3}}{6\sqrt{3}}=11\in N$