Question

$x= \frac{1}{4} \sqrt{48} + \frac{1}{2} \sqrt{243} - \frac{1}{6} \sqrt{12}$

Answer 1

$\dfrac{1}{4} \sqrt{48} + \dfrac{1}{2} \sqrt{243} - \dfrac{1}{6} \sqrt{12}$

Fatore os valore dentro da raiz

48 | 2
24 | 2
12 | 2
  6 | 2
  3 | 3
  1 | ===2⁴ . 3

243 | 3
  81 | 3
  27 | 3
    9 | 3
    3 | 3
    1 |===3⁵

12 | 2
  6 | 2
  3 | 3
  1 |===2² . 3

===
$\dfrac{1}{4} \sqrt{2^4. 3} + \dfrac{1}{2} \sqrt{3^5} - \dfrac{1}{6} \sqrt{2^2. 3} \\ \\ \\ \dfrac{1}{4}. 4\sqrt{3} + \dfrac{1}{2}. 9\sqrt{3} - \dfrac{1}{6}.2 \sqrt{3} \\ \\ \\ \dfrac{ 4\sqrt{3}}{4} + \dfrac{ 9\sqrt{3} }{2}- \dfrac{2 \sqrt{3}}{6} \\ \\ \\ \sqrt{3} + \dfrac{ 9\sqrt{3} }{2}- \dfrac{\sqrt{3}}{3} \\ \\ \\ \sqrt{3} + \dfrac{27\sqrt{3}- 2\sqrt{3} }{6} \\ \\ \\ \sqrt{3} + \dfrac{25\sqrt{3}}{6} \\ \\ \\ \dfrac{6 \sqrt{3} +25\sqrt{3}}{6} \\ \\ \\=\ \textgreater \ \dfrac{31 \sqrt{3} }{6}$