Question

Como racionalizo $\frac{ 8^{3x-4}}{\sqrt{4^{2-x}}}$ ?

Answer 1

Racionalizando:

$\frac{8^{3x - 4}}{\sqrt{4^{2 - x}}} \cdot \frac{\sqrt{4^{2 - x}}}{\sqrt{4^{2 - x}}} = \frac{8^{3x - 4} \cdot \sqrt{4^{2 - x}}}{4^{2 - x}}$

Simplificando:

$\frac{8^{3x - 4} \cdot \sqrt{4^{2 - x}}}{4^{2 - x}} = \frac{(2^{3})^{3x - 4} \cdot [(2^{2})^{2 - x}]^{\frac{1}{2}}}{(2^{2})^{2 - x}} = \frac{2^{9x - 12} \cdot 2^{2 - x}}{2^{4 - 2x}} = \frac{2^{8x - 10}}{2^{4 - 2x}} = \boxed{2^{10x - 14}}$

Answer 2

Só multiplicar o numerador e o denominador por $\sqrt{4^{2-x}}$, isso transformará o denominador em $(\sqrt{4^{2-x}})^{2}$, que é igual a $4^{2-x}$.

$\frac{8^{3x-4}}{\sqrt{4^{2-x}}}=\frac{8^{3x-4}*\sqrt{4^{2-x}}}{\sqrt{4^{2-x}}*\sqrt{4^{2-x}}}$

$\frac{8^{3x-4}}{\sqrt{4^{2-x}}}=\frac{8^{3x-4}*\sqrt{4^{2-x}}}{4^{2-x}}$

Já racionalizamos, mas tem como simplificar ainda (Vou parar de usar a fração porque a letra fica muito pequena)

$8^{3x-4}*\sqrt{4^{2-x}}/(4^{2-x})=(2^{3})^{3x-4}*4^{(2-x)/2}/([2^{2}]^{2-x})$
$8^{3x-4}*\sqrt{4^{2-x}}/(4^{2-x})=2^{9x-12}*(2^{2})^{(2-x)/2}/(2^{4-2x})$
$8^{3x-4}*\sqrt{4^{2-x}}/(4^{2-x})=2^{9x-12}*2^{2-x}/(2^{4-2x})$
$8^{3x-4}*\sqrt{4^{2-x}}/(4^{2-x})=2^{9x-12+2-x}/(2^{4-2x})$
$8^{3x-4}*\sqrt{4^{2-x}}/(4^{2-x})=2^{8x-10}/(2^{4-2x})$
$8^{3x-4}*\sqrt{4^{2-x}}/(4^{2-x})=2^{8x-10-4+2x}$
$8^{3x-4}*\sqrt{4^{2-x}}/(4^{2-x})=2^{10x-14}$

$\boxed{\boxed{8^{3x-4}/\sqrt{4^{2-x}}=2^{10x-14}}}$