Resposta:
a)[tex]\frac{4}{\sqrt{2} +\sqrt{2} } = \frac{4}{2\sqrt{2} } = \frac{2}{\sqrt{2} } . \frac{\sqrt{2} }{\sqrt{2} } =\frac{2\sqrt{2} }{2} =\sqrt{2}[/tex]
b)[tex]\frac{5}{3-\sqrt{2} } .\frac{3+\sqrt{2} }{3+\sqrt{2} } =\frac{5(3+\sqrt{2}) }{3^{2}-\sqrt{2^{2} } }=\frac{15+5\sqrt{2} }{9-2} =\frac{15+5\sqrt{2} }{7}[/tex]
c)[tex]\frac{8}{\sqrt{7} -\sqrt{2} } .\frac{\sqrt{7}+\sqrt{2} }{\sqrt{7}+\sqrt{2} } =\frac{8(\sqrt{7}+\sqrt{2}) }{\sqrt{7^{2} }+\sqrt{2^{2} } } =\frac{8\sqrt{7}+8\sqrt{2} }{7-2} =\frac{8\sqrt{7} +8\sqrt{2} }{5}[/tex]
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Resposta:
a)[tex]\frac{4}{\sqrt{2} +\sqrt{2} } = \frac{4}{2\sqrt{2} } = \frac{2}{\sqrt{2} } . \frac{\sqrt{2} }{\sqrt{2} } =\frac{2\sqrt{2} }{2} =\sqrt{2}[/tex]
b)[tex]\frac{5}{3-\sqrt{2} } .\frac{3+\sqrt{2} }{3+\sqrt{2} } =\frac{5(3+\sqrt{2}) }{3^{2}-\sqrt{2^{2} } }=\frac{15+5\sqrt{2} }{9-2} =\frac{15+5\sqrt{2} }{7}[/tex]
c)[tex]\frac{8}{\sqrt{7} -\sqrt{2} } .\frac{\sqrt{7}+\sqrt{2} }{\sqrt{7}+\sqrt{2} } =\frac{8(\sqrt{7}+\sqrt{2}) }{\sqrt{7^{2} }+\sqrt{2^{2} } } =\frac{8\sqrt{7}+8\sqrt{2} }{7-2} =\frac{8\sqrt{7} +8\sqrt{2} }{5}[/tex]