Question

Ayudenme con el punto 7 y 8 de la última hoja, doy 100 puntos. Completos porfa.

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Answer 1

Respuesta:

Explicación paso a paso:

6-

1) $\frac{4}{\sqrt{6} } =\frac{4}{\sqrt{6} } * \frac{\sqrt{6} }{\sqrt{6} } = \frac{4 * \sqrt{6} }{6} = \frac{2\sqrt{6} }{3}$

2)$\frac{-3}{\sqrt{5} } * \frac{\sqrt{5} }{\sqrt{5} } = \frac{-3*\sqrt{5} }{5}$

3) $\frac{5}{\sqrt{10} } *\frac{\sqrt{10} }{\sqrt{10} } =\frac{5*\sqrt{10} }{10} =\frac{\sqrt{10} }{2}$

4)$\frac{\sqrt{7} }{\sqrt{2}} *\frac{\sqrt{2} }{\sqrt{2} } =\frac{\sqrt{14} }{2}$

5) $\frac{\sqrt{12} }{\sqrt{5} } *\frac{\sqrt{5} }{\sqrt{5} } = \frac{2*\sqrt{15} }{5}$

6)$\frac{\sqrt{3} }{\sqrt{6} } *\frac{\sqrt{6} }{\sqrt{6} } = \frac{3*\sqrt{2} }{6} =\frac{\sqrt{2} }{2}$

7)$\frac{1+\sqrt{3} }{\sqrt{3} } *\frac{\sqrt{3} }{\sqrt{3} } =\frac{\sqrt{3}+3 }{3} = \frac{\sqrt{3} }{3} +1$

8)$\frac{5-\sqrt{2} }{\sqrt{5} }*\frac{\sqrt{5} }{\sqrt{5} } =\frac{5*\sqrt{5} -\sqrt{10} }{5} = \sqrt{5} -\frac{\sqrt{10} }{5}$

9)$\frac{\sqrt{3}- \sqrt{6} }{\sqrt{3} } *\frac{\sqrt{3} }{\sqrt{3} } = \frac{3 - \sqrt{18} }{3} = \frac{3-3*\sqrt{2} }{3} = 1 -\sqrt{2}$

7-

a)$\frac{1}{\sqrt[5]{x^{2} } } *\frac{\sqrt[5]{x^{3} } }{\sqrt[5]{x^{3} } } =\frac{\sqrt[5]{x^{3} } }{x}$

b) $\frac{11}{2\sqrt{3} -1} *\frac{2\sqrt{3}+1 }{2\sqrt{3}+1} =\frac{22\sqrt{3} +11 }{4*3-1^{2} }=\frac{22\sqrt{3}+11 }{11} = 2\sqrt{3} +1$

c)$\frac{3-\sqrt{2} }{1 + \sqrt{2} } *\frac{1-\sqrt{2} }{1-\sqrt{2}} = \frac{3-3\sqrt{2}-\sqrt{2} +2 }{1-2} = \frac{5 - 4\sqrt{2} }{-1} = 4\sqrt{2} - 5$

d) $\frac{6}{\sqrt[4]{3} } *\frac{\sqrt[4]{3^{3} } }{\sqrt[4]{3^{3} } } = \frac{6 * \sqrt[4]{3^{3} } }{3} =2\sqrt[4]{3^{3} }$

e)$\frac{12}{\sqrt{5}+\sqrt{2} } *\frac{\sqrt{5}-\sqrt{2} }{\sqrt{5} -\sqrt{2} } =\frac{12\sqrt{5} -12\sqrt{2} }{5-2} =4\sqrt{5} -4\sqrt{2}$

f)$\frac{5-2\sqrt{2} }{1-\sqrt{2} } *\frac{1+\sqrt{2} }{1+\sqrt{2} } = \frac{5+5\sqrt{2} -2\sqrt{2} -4}{-1} =-1-3\sqrt{2}$

g) $\frac{3}{\sqrt[3]{9x} } * \frac{\sqrt[3]{3x^{2} } }{\sqrt[3]{3x^{2} } } =\frac{3\sqrt[3]{3x^{2} } }{3x} = \frac{\sqrt[3]{3x^{2} } }{x}$

h) $\frac{2\sqrt{7} }{\sqrt{7} -2} *\frac{\sqrt{7}+2 }{\sqrt{7} +2} =\frac{14+4\sqrt{7} }{7-4} = \frac{14}{3} + \frac{4\sqrt{7} }{3}$

i) $\frac{3 - \sqrt{6} }{3 + \sqrt{6 } } * \frac{3- \sqrt{6} }{3 -\sqrt{6} } = \frac{9 +6 - 6 \sqrt{6} }{9 - 6} = \frac{15 - 6 \sqrt{6} }{3} = 5 - 2\sqrt{6}$