Question

Racionalizar : raiz de 3 - raiz de 1 sobre raiz de 4 + raiz de 12
Ayueenmee porfavorr

Answer 1

Respuesta:

$\frac{\sqrt{3}-1}{\sqrt{4}+\sqrt{12}}=\frac{8-4\sqrt{3}}{8}$

Explicación paso a paso:

$\frac{\sqrt{3}-1}{\sqrt{4}+\sqrt{12}}\\\textbf{Multiplicando la fracci\'on}\\\textbf{por el conjugado del denominador}\\\\\left(\frac{\sqrt{4}-\sqrt{12}}{\sqrt{4}-\sqrt{12}}\right)\left(\frac{\sqrt{3}-1}{\sqrt{4}+\sqrt{12}}\right)=\left(\frac{(\sqrt{4}-\sqrt{12})(\sqrt{3}-1)}{\left(\sqrt{4}\right)^{2}-\left(\sqrt{12}\right)^{2}}\right)\,\,((a+b)(a-b)=a^{2}-b^{2})\\\\\textbf{Desarrollando y simplificando}\\\frac{(\sqrt{4}-\sqrt{12})(\sqrt{3}-1)}{4-12}=\frac{-\sqrt{4}+\sqrt{12}+\sqrt{3}\sqrt{4}-\sqrt{3}\sqrt{12}}{-8}=\frac{-2+2\sqrt{3}+2\sqrt{3}-2\sqrt{3}\sqrt{3}}{-8}=\\\\=\frac{-2+2\sqrt{3}+2\sqrt{3}-2(\sqrt{3})^{2}}{-8}=\frac{-2+4\sqrt{3}-2(3)}{-8}=\\\\=\frac{-2+4\sqrt{3}-6}{-8}=\frac{-8+4\sqrt{3}}{-8}=\\\\=\frac{8-4\sqrt{3}}{8}$