Question

qual e o valor de x nas expressõ
$\frac{1}{4} \sqrt{48} + \frac{1}{2} \sqrt{243} - \frac{1}{6 } \sqrt{12}$
​

Answer 1

Resposta:

$\frac{31}{6}\sqrt{3}$

Explicação passo-a-passo:

Primeiro decompõe os número dentro das raízes:

$48=2^{4}.3\\243=3^{5}\\12=2^{2}.3$

$\frac{1}{4}\sqrt{48}+\frac{1}{2}\sqrt{243}-\frac{1}{6}\sqrt{12}=\frac{1}{4}\sqrt{2^{4}.3}+\frac{1}{2} \sqrt{3^{5}}-\frac{1}{6}\sqrt{2^2.3}=\\\\\frac{1}{4}(2^{4}.3)^\frac{1}{2}+\frac{1}{2}(3^{5})^\frac{1}{2}-\frac{1}{6}(2^2.3)^\frac{1}{2}=\\\\\frac{1}{4}(2^{\frac{4.1}{2}}.3^\frac{1}{2})+\frac{1}{2}(3)^\frac{5.1}{2}-\frac{1}{6}.2^\frac{2.1}{2}.3^\frac{1}{2}=$

$\frac{1}{4}(2^{2}.3^\frac{1}{2})+\frac{1}{2}(3^\frac{4}{2}3^{\frac{1}{2}})-\frac{1}{6}(2.3^\frac{1}{2})=\\\\3^\frac{1}{2}.[\frac{1}{4}(2^{2})+\frac{1}{2}(3^{2})-\frac{1}{6}(2)]=\\\\3^\frac{1}{2}.[\frac{4}{4}+\frac{9}{2}-\frac{2}{6}]=\\\\3^\frac{1}{2}.[1+\frac{9}{2}-\frac{1}{3}]=\\\\3^\frac{1}{2}.[\frac{6.1+3.9-2.1}{6}]=\\\\3^\frac{1}{2}.[\frac{31}{6}]=\\\\\frac{31}{6}\sqrt{3}$

Propriedades:

$(a^{m})^{n}=a^{m.n}\\\\\sqrt[n]{x^m} =x^{\frac{m}{n} }\\\\a^{m}a^{n}=a^{m+n}$