Question

No caderno,racionalize o denominador de cada expressão a seguir : (gente me ajudem não to conseguindo) : a)1- √3 sobre raiz quadrada de 3 = b)3- √2 sobre raiz quadrada de 2 = c) √5+ √2 sobre raiz quadrada de 5 =

Answer 1

$<var>A)\frac{1-\sqrt3}{\sqrt3}=>\frac{(1-\sqrt3)\sqrt3}{\sqrt3*\sqrt3}=>\frac{\sqrt3-3}{3}\\\\B)\frac{3-\sqrt2}{\sqrt2}=>\frac{(3-\sqrt2)\sqrt2}{\sqrt2*\sqrt2}=>\frac{3\sqrt2-2}{2}\\\\C)\frac{\sqrt5+\sqrt2}{\sqrt5}=>\frac{(\sqrt5+\sqrt2)\sqrt5}{\sqrt5*\sqrt5}=>\frac{5+\sqrt10}{5}</var>$

Answer 2

1) $\dfrac{1-\sqrt{3}}{\sqrt{3}}$

 

Temos que:

 

$\dfrac{1-\sqrt{3}}{\sqrt{3}}=\dfrac{1-\sqrt{3}}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{\sqrt{3}-3}{3}$

 

 

2) $\dfrac{3-\sqrt{2}}{\sqrt{2}}$

 

Analogamente, temos:

 

$\dfrac{3-\sqrt{2}}{\sqrt{2}}=\dfrac{3-\sqrt{2}}{\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}=\dfrac{3\sqrt{2}-2}{2}$

 

 

3) $\dfrac{\sqrt{5}+\sqrt{2}}{\sqrt{5}}$

 

Temos que:

 

$\dfrac{\sqrt{5}+\sqrt{2}}{\sqrt{5}}=\dfrac{\sqrt{5}+\sqrt{2}}{\sqrt{5}}\cdot\dfrac{\sqrt{5}}{\sqrt{5}}=\dfrac{5+\sqrt{10}}{5}$