Question

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Answer 1

Răspuns:

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Answer 2

$\it A=\sqrt{32} -\sqrt{8} +\sqrt{98} +\sqrt{128} +\sqrt{18} =\sqrt{16\cdot2}-\sqrt{4\cdot2} +\sqrt{49\cdot 2}-\sqrt{64\cdot 2}+\\ \\ +\sqrt{9\cdot2} =4\sqrt2-2\sqrt2+7\sqrt2-8\sqrt2+3\sqrt2=14\sqrt2-10\sqrt2=4\sqrt2\\ \\ \\ B=\sqrt{75} -\sqrt{147} +\sqrt{243} +\sqrt{108} +\sqrt{12} =\sqrt{25\cdot3}-\sqrt{49\cdot3} +\sqrt{81\cdot 3}-\\ \\ -\sqrt{36\cdot 3}+\sqrt{4\cdot3} =5\sqrt3 -7\sqrt3+9\sqrt3-6\sqrt3+2\sqrt3=3\sqrt3$

$\it C=\sqrt{54} -\sqrt{384} +\sqrt{96} +\sqrt{24} +\sqrt{150} = \sqrt{9\cdot6}-\sqrt{64\cdot6} +\sqrt{16\cdot 6}-\sqrt{4\cdot 6}+ \\ \\ + \sqrt{25\cdot6}=3\sqrt6 -8\sqrt6+4\sqrt6-2\sqrt6+5\sqrt6 =12\sqrt6 -10\sqrt6 =2\sqrt6$

$\it \dfrac{A\cdot B}{C} =\dfrac{4\sqrt2\cdot 3\sqrt3}{2\sqrt6} =\dfrac{12}{2} = 6 \\ \\ \\ \dfrac{A\cdot C}{B} =\dfrac{4\sqrt2\cdot 2\sqrt6}{3\sqrt3} = \dfrac{4\cdot 2\cdot 2}{3} =\dfrac{16}{3}$

$\it \dfrac{B\cdot C}{A} =\dfrac{3\sqrt3\cdot2\sqrt6}{4\sqrt2} =\dfrac{3\sqrt3\cdot\sqrt3}{2} =\dfrac{9}{2} =4,5$