Question

Racionalização de denominadores

Answer 1

Quando temos uma fração do tipo:

$\dfrac{numerador}{\sqrt{a}\pm\sqrt{b}}$

Racionalizamos dessa maneira:

$\boxed{\boxed{\dfrac{numerador}{\sqrt{a}\pm\sqrt{b}}\cdot\dfrac{\sqrt{a}\mp\sqrt{b}}{\sqrt{a}\mp\sqrt{b}}}}$

Produto da soma pela diferença de dois termos:

$\boxed{\boxed{(x+y)\cdot(x-y)=x^{2}-y^{2}}}$
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$\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\dfrac{(\sqrt{3}-\sqrt{2})(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}\\\\\\\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\dfrac{(\sqrt{3}-\sqrt{2})^{2}}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}\\\\\\\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\dfrac{(\sqrt{3})^{2}-2\cdot\sqrt{3}\cdot\sqrt{2}+(\sqrt{2})^{2}}{3-2}\\\\\\\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\dfrac{3-2\sqrt{3\cdot2}+2}{1}$

Portanto:

$\boxed{\boxed{\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}=5-2\sqrt{6}}}$

Answer 2

(√3 - √2) .   (√3 - √2)    =  (√3)² - √2√3 - √3.√2 + √2√2) = 3 - 2√6 + 2
(√3 + √2)      (√3 - √2)       (√3)² - √2√3 + √2√3 - 2          3 - 2

= 3 - 2√6 +2 = 5 - 2√6