Question

$\frac{1}{2} \sqrt[3]{8} {x}^{3} {y}^{3}$
$2 \sqrt{4} {x}^{2}$
$\frac{1}{4} \sqrt[4]{81} {m}^{8} {m}^{4}$
$\sqrt[4]{32} {a}^{9} {b}^{4}$
​

Answer 1

Respuesta:

Facil...

Explicación paso a paso:

1)

$\frac{1}{2}\sqrt[3]{8x^3y^3}\\\\=2\cdot \frac{1}{2}xy\\\\\mathrm{Multiplicar\:fracciones}:\quad \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c}\\\\=\frac{1\cdot \:2xy}{2}\\\\=1\cdot \:xy\\\\=xy$

2)

$2\sqrt{4x^2}\\\\=2\cdot \:2x\\\\=4x$

3)

$\frac{1}{4}\sqrt[4]{81m^8m^4}\\\\=3\cdot \frac{1}{4}m^3\\\\=\frac{1\cdot \:3m^3}{4}\\\\=\frac{3m^3}{4}$

4)

$\sqrt[4]{32a^9b^4}\\\\=\sqrt[4]{32}\sqrt[4]{a^9}\sqrt[4]{b^4}\\\\=\sqrt[4]{32}a^2\sqrt[4]{a}\sqrt[4]{b^4}\\\\\mathrm{Aplicar\:la\:siguiente\:propiedad\:de\:los\:radicales:}\:\sqrt[n]{a^n}=a.\\\\=\sqrt[4]{32}a^2\sqrt[4]{a}b\\\\=2\sqrt[4]{2}a^2\sqrt[4]{a}b\\\\=2\sqrt[4]{2}a^2b\sqrt[4]{a}$