Question

racionalize as expressões abaixo :

a) 2/ ✓5 + ✓3
b) 11/ 2✓3 - 1

Answer 1

Lembrete:

๏ $\left(a+b\right)\cdot \left(a-b\right)=a^2-b^2$
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a) 

$\dfrac{2}{\sqrt{5}+\sqrt{3}}=\\\\\dfrac{2}{\sqrt{5}+\sqrt{3}}\cdot \dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}=\\\\\dfrac{2\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\cdot \sqrt{5}-\sqrt{3}}=\\\\\dfrac{2\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}\right)^2-\left(\sqrt{3}\right)^2}=\\\\\dfrac{2\left(\sqrt{5}-\sqrt{3}\right)}{5-3}=\\\dfrac{2\left(\sqrt{5}-\sqrt{3}\right)}{2}=\\\\\boxed{\bold{\sqrt{5}-\sqrt{3}}}$

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b)

$\dfrac{11}{2\sqrt{3}-1}=\\\\\dfrac{11}{2\sqrt{3}-1}\cdot \dfrac{2\sqrt{3}+1}{2\sqrt{3}+1}=\\\\\dfrac{11\cdot \left(2\sqrt{3}+1\right)}{\left(2\sqrt{3}-1\right)\cdot 2\sqrt{3}+1}=\\\\\dfrac{11\cdot \left(2\sqrt{3}+1\right)}{\left(2\sqrt{3}\right)^2-1^2}=\\\\\dfrac{11\cdot \left(2\sqrt{3}+1\right)}{\left(4\cdot 3\right)-1}=\\\\\dfrac{11\cdot \left(2\sqrt{3}+1\right)}{12-1}=\\\\\dfrac{11\cdot \left(2\sqrt{3}+1\right)}{11}=\\\\\boxed{\bold{2\sqrt{3}+1}}$