Question

Simplificando a expressão

√3 / √3 - 1 - 1 / √3 +
2 / 1 + √3

Vamos obter:

a)7 √3 + 3

b)-5 √3 + 15 / 6

c)5 √3 - 15 / 6

d)7 √3 + 3 / 6

e)7 √3 + 1 / 2

Urgente, me ajudem!​

Answer 1

Queremos simplificar a seguinte expressão:

$\mathsf{\dfrac{\sqrt{3}}{\sqrt{3}-1}-\dfrac{1}{\sqrt{3}}+\dfrac{\sqrt{2}}{1+\sqrt{3}}}$

De início, vamos racionalizar o denominador de cada parcela. Veja:

$\mathsf{\dfrac{\sqrt{3}}{\sqrt{3}-1}\cdot {\color{blue}\dfrac{\sqrt{3}+1}{\sqrt{3}+1}}-\dfrac{1}{\sqrt{3}}\cdot{\color{blue}\dfrac{\sqrt{3}}{\sqrt{3}}}+\dfrac{\sqrt{2}}{1+\sqrt{3}}\cdot {\color{blue}\dfrac{1-\sqrt{3}}{1-\sqrt{3}}}}$

Note que multiplicamos o numerador e o denominador de cada fração pelo fator racionalizante de cada denominador.

Agora vamos simplicar a expressão:

$\mathsf{\dfrac{\sqrt{3}}{\sqrt{3}-1}\cdot {\color{blue}\dfrac{\sqrt{3}+1}{\sqrt{3}+1}}-\dfrac{1}{\sqrt{3}}\cdot{\color{blue}\dfrac{\sqrt{3}}{\sqrt{3}}}+\dfrac{\sqrt{2}}{1+\sqrt{3}}\cdot {\color{blue}\dfrac{1-\sqrt{3}}{1-\sqrt{3}}}}=\\=\mathsf{\dfrac{3+\sqrt{3}}{\left(\sqrt{3}\right)^2-1^2}-\dfrac{\sqrt{3}}{\left(\sqrt{3}\right)^2}}+\dfrac{2-2\sqrt{3}}{1^2-\left(\sqrt{3}\right)^2}=\\=\mathsf{\dfrac{3+\sqrt{3}}{3-1}+\dfrac{\sqrt{3}}{3}+\dfrac{2-2\sqrt{3}}{1-3}}=\\=\mathsf{\dfrac{3+\sqrt{3}}{2}+\dfrac{\sqrt{3}}{3}+\dfrac{2-2\sqrt{3}}{-2}}=\\=\mathsf{\dfrac{3+\sqrt{3}}{2}+\dfrac{\sqrt{3}}{3}+\dfrac{-2+2\sqrt{3}}{2}}$

Reduzindo cada parcela ao mesmo denominador, temos:

$\mathsf{\dfrac{3+\sqrt{3}}{2}+\dfrac{\sqrt{3}}{3}+\dfrac{-2+2\sqrt{3}}{2}}=\\=\mathsf{\dfrac{3(3+\sqrt{3})-2\sqrt{3}+3(-2+2\sqrt{3})}{6}}=\\=\mathsf{\dfrac{9+3\sqrt{3}-2\sqrt{3}-6+6\sqrt{3}}{6}}=\\=\mathsf{\dfrac{3+7\sqrt{3}}{6}}=\\=\mathsf{\dfrac{7\sqrt{3}+3}{6}}$

Portanto,

$\fbox{\fbox{\mathsf{\dfrac{\sqrt{3}}{\sqrt{3}-1}-\dfrac{1}{\sqrt{3}}+\dfrac{\sqrt{2}}{1+\sqrt{3}}=\dfrac{7\sqrt{3}+3}{6}}}}$

Resposta: alternativa D.