Question

Va rog ajutati ma la subiectul al 2 lea ex 1​

Answer 1

$\it \sqrt{12} =\sqrt{4\cdot3}=2\sqrt3\\ \\ \sqrt{45}=\sqrt{25\cdot3}=5\sqrt3\\ \\ \sqrt{48}=\sqrt{16\cdot3}=4\sqrt3\\ \\ \sqrt{27}=\sqrt{9\cdot3}=3\sqrt3$

Exercițiul devine:

$\it \Big(\dfrac{3}{2\sqrt3}-\dfrac{2}{5\sqrt3}+\dfrac{3}{8\sqrt3}-\dfrac{21^{(3}}{5\cdot3\sqrt3}\Big)\cdot\dfrac{4}{\sqrt6}=\Big(\dfrac{^{20)}3}{\ 2}-\dfrac{^{8)}2}{\ 5}+\dfrac{^{5)}3}{\ 8}-\dfrac{^8)7}{\ 5}\Big)\cdot\dfrac{1}{\sqrt3}\cdot\dfrac{4}{\sqrt6}=\\ \\ \\ =\dfrac{60-16+15-56}{40}\cdot\dfrac{4}{\sqrt{18}}=\dfrac{\not3}{\not4\cdot10}\cdot\dfrac{\not{4}}{\not3\sqrt2}=\dfrac{^{\sqrt2)}1}{10\sqrt2}=\dfrac{\sqrt2}{20}$