Verified answer

Resposta:

1.
a) [tex]\frac{2\sqrt{10} }{10}[/tex].

b) [tex]\sqrt{6}[/tex].

c) [tex]3\sqrt{3}[/tex].

d) [tex]\frac{2\sqrt{5} }{5}[/tex].

e) [tex]\frac{20\sqrt{3} }{3}[/tex].

f) [tex]\frac{2\sqrt{10} }{10}[/tex].

2)
a) [tex]\frac{-2\sqrt{3} }{3}[/tex].

b) [tex]3[/tex].

c) [tex]\frac{\sqrt{15} + \sqrt{6} }{3}[/tex].

d) [tex]\frac{3(\sqrt{3} + \sqrt{2}) }{\sqrt{9} - 2 }[/tex].

e) [tex]\frac{2(\sqrt{2} + 1) }{2}[/tex].

f) [tex]\frac{\sqrt{5} + \sqrt{10} }{5}[/tex].

3)
a) [tex]\frac{3 + \sqrt{6} }{3}[/tex].

b) [tex]\frac{2(\sqrt{5} - \sqrt{3}) }{5 - \sqrt{9} }[/tex].

c) [tex]\frac{8 - 5\sqrt{2} }{7}[/tex].

d) [tex]- 3 + 2(\sqrt{3} - \sqrt{2}) + \sqrt{6}[/tex].

4)
a) [tex]\frac{\sqrt[5]{36} }{6}[/tex].

b) [tex]\sqrt[9]{4}[/tex].

c) [tex]\frac{\sqrt[4]{8} }{2}[/tex].

d) [tex]2\sqrt[11]{10^{6} }[/tex]

Explicação passo a passo:
O objetivo da racionalização de denominadores é eliminar o radical no denominador da fração. Para isso, multiplicamos tanto o denominador quanto o numerador por outra radical que faça o resultado ser um número natural.

1.
a) [tex]\frac{2}{\sqrt{10} } . \frac{\sqrt{10} }{\sqrt{10} } = \frac{2\sqrt{10} }{\sqrt{100} }[/tex]
[tex]= \frac{2\sqrt{10} }{10}[/tex]

b) [tex]\frac{6}{\sqrt{6} } . \frac{\sqrt{6} }{\sqrt{6} } = \frac{6\sqrt{6} }{\sqrt{36} }[/tex]
[tex]= \frac{6\sqrt{6} }{6} = \sqrt{6}[/tex]

c) [tex]\frac{9}{\sqrt{3} } . \frac{\sqrt{3} }{\sqrt{3} } = \frac{9\sqrt{3} }{\sqrt{9} }[/tex]
[tex]= \frac{9\sqrt{3} }{3} = 3\sqrt{3}[/tex]

d) [tex]\frac{2}{\sqrt{5} } . \frac{\sqrt{5} }{\sqrt{5} } = \frac{2\sqrt{5} }{\sqrt{25} }[/tex]
[tex]= \frac{2\sqrt{5} }{5}[/tex]

e) [tex]\frac{20}{\sqrt{3} } . \frac{\sqrt{3} }{\sqrt{3} } = \frac{20\sqrt{3} }{\sqrt{9} }[/tex]
[tex]= \frac{20\sqrt{3} }{3}[/tex]

f) [tex]\frac{2\sqrt{5} }{5\sqrt{2} } . \frac{\sqrt{2} }{\sqrt{2} } = \frac{2\sqrt{10} }{5\sqrt{4} }[/tex]

[tex]= \frac{2\sqrt{10} }{5 . 2} = \frac{2\sqrt{10} }{10}[/tex]

2)
a) [tex]\frac{1 - 3}{\sqrt{3} } . \frac{\sqrt{3} }{\sqrt{3} } = \frac{\sqrt{3} - 3\sqrt{3} }{\sqrt{9} }[/tex]

[tex]= \frac{\sqrt{3} - 3\sqrt{3} }{3} = \frac{-2\sqrt{3} }{3}[/tex]

b) [tex]\frac{3\sqrt{2} }{\sqrt{2} } . \frac{\sqrt{2} }{\sqrt{2} } = \frac{3\sqrt{4} }{\sqrt{4} }[/tex]
[tex]= \frac{3 . 2}{2} = \frac{6}{2}[/tex]
[tex]= 3[/tex]

c) [tex]\frac{\sqrt{5} + \sqrt{2} }{\sqrt{3} } . \frac{\sqrt{3} }{\sqrt{3} } = \frac{\sqrt{15} + \sqrt{6} }{\sqrt{9} }[/tex]
[tex]= \frac{\sqrt{15} + \sqrt{6} }{3}[/tex]

d) [tex]\frac{3}{\sqrt{3} - \sqrt{2} } . \frac{\sqrt{3} + \sqrt{2} }{\sqrt{3} + \sqrt{2} } = \frac{3\sqrt{3} + 3\sqrt{2} }{\sqrt{9 + \sqrt{6} - \sqrt{6} - \sqrt{4} } }[/tex]
[tex]= \frac{3\sqrt{3} + 3\sqrt{2} }{\sqrt{9} - \sqrt{4} } = \frac{3(\sqrt{3} + \sqrt{2}) }{\sqrt{9} - 2 }[/tex]

e) [tex]\frac{2 + \sqrt{2} }{\sqrt{2} } . \frac{\sqrt{2} }{\sqrt{2} } = \frac{2\sqrt{2} + \sqrt{4} }{\sqrt{4} }[/tex]

[tex]= \frac{2\sqrt{2} + 2 }{2} = \frac{2(\sqrt{2} + 1) }{2}[/tex]

f) [tex]\frac{1 + \sqrt{2} }{\sqrt{5} } . \frac{\sqrt{5} }{\sqrt{5} } = \frac{\sqrt{5} + \sqrt{10} }{\sqrt{25} }[/tex]
[tex]= \frac{\sqrt{5} + \sqrt{10} }{5}[/tex]

3)
a) [tex]\frac{1}{3 - \sqrt{6} } . \frac{3 + \sqrt{6} }{3 + \sqrt{6} } = \frac{3 + \sqrt{6} }{9 + 3\sqrt{6} - 3\sqrt{6} - \sqrt{36} }[/tex]
[tex]= \frac{3 + \sqrt{6} }{9 - \sqrt{36} } = \frac{3 + \sqrt{6} }{9 - 6}[/tex]
[tex]= \frac{3 + \sqrt{6} }{3}[/tex]

b) [tex]\frac{2}{\sqrt{5} + \sqrt{3} } . \frac{\sqrt{5} - \sqrt{3} }{\sqrt{5} - \sqrt{3} } = \frac{2\sqrt{5} - 2\sqrt{3} }{\sqrt{25} - \sqrt{15} + \sqrt{15} - \sqrt{9} }[/tex]
[tex]\frac{2\sqrt{5} - 2\sqrt{3} }{5 - \sqrt{9} } = \frac{2(\sqrt{5} - \sqrt{3}) }{5 - \sqrt{9} }[/tex]

c) [tex]\frac{2 - \sqrt{2} }{3 + \sqrt{2} } . \frac{3 - \sqrt{2} }{3 - \sqrt{2} } = \frac{6 - 2\sqrt{2} - 3\sqrt{2} + \sqrt{4} }{9 - 3\sqrt{2} + 3\sqrt{2} - \sqrt{4} }[/tex]
[tex]= \frac{6 - 5\sqrt{2} + 2}{9 - 2} = \frac{8 - 5\sqrt{2} }{7}[/tex]

d) [tex]\frac{\sqrt{3} - \sqrt{2} }{\sqrt{3} + 2} . \frac{\sqrt{3} - 2 }{\sqrt{3 - 2} } = \frac{\sqrt{9} - 2\sqrt{3} - \sqrt{6} + 2\sqrt{2} }{\sqrt{9} - 2\sqrt{3} + 2\sqrt{3} - 4 }[/tex]

[tex]= \frac{3 -2\sqrt{3} - \sqrt{6} + 2\sqrt{2} }{3 - 4} = \frac{3 -2\sqrt{3} - \sqrt{6} + 2\sqrt{2} }{- 1}[/tex]
[tex]= - 3 + 2\sqrt{3} + \sqrt{6} - 2\sqrt{2} = - 3 + 2(\sqrt{3} - \sqrt{2}) + \sqrt{6}[/tex]

4)
a) [tex]\frac{1}{\sqrt[5]{6^{3} } } . \frac{\sqrt[5]{6^{2} } }{\sqrt[5]{6^{2} } } = \frac{\sqrt[5]{6^{2} } }{\sqrt[5]{6^{5} } }[/tex]
[tex]= \frac{\sqrt[5]{36} }{6}[/tex]

b) [tex]\frac{2}{\sqrt[9]{2^{7} } } . \frac{\sqrt[9]{2^{2} } }{\sqrt[9]{2^{2} } } = \frac{2\sqrt[9]{2^{2} } }{\sqrt[9]{2^{9} } }[/tex]
[tex]= \frac{2\sqrt[9]{4} }{2} = \sqrt[9]{4}[/tex]

c) [tex]\frac{4}{\sqrt[4]{8^{3} } } . \frac{\sqrt[4]{8} }{\sqrt[4]{8} } = \frac{4\sqrt[4]{8} }{\sqrt[4]{8^{4} } }[/tex]
[tex]= \frac{4\sqrt[4]{8} }{8} = \frac{\sqrt[4]{8} }{2}[/tex]

d) [tex]\frac{20}{\sqrt[11]{10^{5} } } . \frac{\sqrt[11]{10^{6} } }{\sqrt[11]{10^{6} } } = \frac{20\sqrt[11]{10^{6} } }{\sqrt[11]{10^{11} } }[/tex]
[tex]\frac{20\sqrt[11]{10^{6} } }{10} = 2\sqrt[11]{10^{6} }[/tex]