Question

am nevoie urgent de aceste ex și rezolvarea va rog ofer coroana
este în poza​

Answer 1

4. media geometrică a două numere este radical din produsul lor

Vom scrie mai întâi cele două numere într-o formă mai simplă, efectuând toate calculele posibile.

$\displaystyle a = \sqrt{4\cdot3} -2\sqrt{9\cdot3} +5\sqrt{3} +\sqrt{16\cdot3} +2\sqrt{25\cdot3}$

$\displaystyle a = 2\sqrt{3} -2\cdot3\sqrt{3} +5\sqrt{3} +4\sqrt{3} +2\cdot5\sqrt{3}$

$\displaystyle a = 2\sqrt{3} -6\sqrt{3} +9\sqrt{3} +10\sqrt{3}$

$\displaystyle a = 15\sqrt{3}$

$\displaystyle b = |5-\sqrt{3}| -|2\sqrt{3}-3\sqrt{2}|-|5-3\sqrt{2}|+\sqrt{49-1}$

$\displaystyle b = 5-\sqrt{3} -(3\sqrt{2}-2\sqrt{3})-(5-3\sqrt{2})+\sqrt{48}$

$\displaystyle b = 5-\sqrt{3} -3\sqrt{2}+2\sqrt{3}-5+3\sqrt{2}+\sqrt{16\cdot3}$

$\displaystyle b = \sqrt{3} +4\sqrt{3}$

$\displaystyle b = 5\sqrt{3}$

Calculăm media geometrică:

$\displaystyle m_{g} =\sqrt{a\cdot b}$

$\displaystyle m_{g} =\sqrt{15\sqrt{3}\cdot 5\sqrt{3} } =\sqrt{15\cdot5\cdot3}$

$\displaystyle \bold {m_{g} =15}$

5.

$\displaystyle x=\sqrt{3} \cdot\left (\frac{\sqrt{3}\cdot \sqrt{2} }{3\sqrt{2}} +\frac{25}{2\cdot\sqrt{25} \cdot\sqrt{3}} \right)-\left (\frac{12}{5\sqrt{3}} -\frac{\sqrt{3}}{2} \right)\cdot\frac{27}{\sqrt{27} } -\left (\frac{\sqrt{16} \cdot\sqrt{3} }{5} +2\sqrt{3} \right)\cdot\frac{1}{\sqrt{3} }$

$\displaystyle x=\sqrt{3} \cdot\left (\frac{\sqrt{3}}{3} +\frac{5}{2\sqrt{3}} \right)-\left (\frac{4\cdot\sqrt{3} }{5} -\frac{\sqrt{3}}{2} \right)\cdot\sqrt{27} -\left (\frac{4\sqrt{3} }{5} +2\sqrt{3} \right)\cdot\frac{1}{\sqrt{3} }$

$\displaystyle x=\sqrt{3} \cdot\frac{2\sqrt{3}+5\sqrt{3} }{2\cdot3} -\frac{8\sqrt{3} -5\sqrt{3}}{10}\cdot3\sqrt{3} -\frac{4\sqrt{3}+10\sqrt{3}}{5} \cdot\frac{1}{\sqrt{3} }$

$\displaystyle x=\frac{\sqrt{3} \cdot\sqrt{3}\cdot7}{2\cdot3} -\frac{3\sqrt{3} \cdot3\sqrt{3}}{10} -\frac{14\sqrt{3}}{5\cdot\sqrt{3}}$

$\displaystyle x=\frac{7}{2} -\frac{27}{10} -\frac{14}{5}$

$\displaystyle x=\frac{35-27-28}{10}$

$\displaystyle x=\frac{-20}{10}$

$\displaystyle \bold {x=-2}$