Question

con el proceso por fa

$\frac{-\sqrt{3} }{2\sqrt{3}-1 }$

Answer 1

RESOLVER:

$\frac{-\sqrt{3}}{2\sqrt{3}-1}$

Debemos de racionalizar el denominador, para ello multiplicamos la operación por el conjugado:  $\frac{2\sqrt{3}+1}{2\sqrt{3}+1}$

$\frac{-\sqrt{3}}{2\sqrt{3}-1}\cdot\frac{2\sqrt{3}+1}{2\sqrt{3}+1}\\\\=\frac{-\sqrt{3}\left(2\sqrt{3}+1\right)}{\left(2\sqrt{3}-1\right)\left(2\sqrt{3}+1\right)}$

Resolvemos el numerador.

$-\sqrt{3}\left(2\sqrt{3}+1\right)\\\\\sqrt{3}\left(2\sqrt{3}+1\right)=\sqrt{3}\cdot \:2\sqrt{3}+\sqrt{3}\cdot \:1\\\\=-\left(\sqrt{3}\cdot \:2\sqrt{3}+\sqrt{3}\cdot \:1\right)$

Multiplicamos:

$\sqrt{3}\cdot \:2\sqrt{3}\\\\\sqrt{3}\sqrt{3}=\sqrt{3^2} =3\\\\=3\cdot \:2\\\\3\cdot \:2=6\\\\=6$

Multiplicamos:

$\sqrt{3}\cdot \:1\\\\=\sqrt{3}$

Ahora resolvemos.

$=-\left(6+\sqrt{3}\right)\\\\-\left(6+\sqrt{3}\right)=-6-\sqrt{3}\\\\=-6-\sqrt{3}$

$=\frac{-6-\sqrt{3}}{\left(2\sqrt{3}-1\right)\left(2\sqrt{3}+1\right)}$

Resolvemos el denominador.

$\left(2\sqrt{3}-1\right)\left(2\sqrt{3}+1\right)\\\\\left(2\sqrt{3}-1\right)\left(2\sqrt{3}+1\right)=\left(2\sqrt{3}\right)^2-1^2\\\\=\left(2\sqrt{3}\right)^2-1^2$

Hallamos la raíz.

$\left(2\sqrt{3}\right)^2\\\\\left(2\sqrt{3}\right)^2=2^2\left(\sqrt{3}\right)^2\\\\=2^2\left(\sqrt{3}\right)^2\\\\=4\left(\sqrt{3}\right)^2\\\\=4\left(\sqrt{3^2}\right)\\\\=4\cdot \:3\\\\=12$

Resolvemos:

$12-1^2\\\\1^2=1\\\\12-1=11\\\\=11$

$\boxed{=\frac{-6-\sqrt{3}}{11}}$

Resultado del numerador:

$-\sqrt{3}\left(2\sqrt{3}+1\right)\Longrightarrow-6-\sqrt{3}$

Resultado del denominador:

$\left(2\sqrt{3}-1\right)\left(2\sqrt{3}+1\right)\Longrightarrow11$

MUCHA SUERTE...