Como racionalizar os denominadores da fração?
18 fev 2015, 14:36
Como racionalizar a seguinte fração?
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Re: Como racionalizar os denominadores da fração? [resolvida]
18 fev 2015, 16:57
Boa tarde!
\(\frac{1}{\sqrt{3}+\sqrt{5}+\sqrt{7}}=\frac{1}{\sqrt{3}+\sqrt{5}+\sqrt{7}}\times \frac{\sqrt{3}+\sqrt{5}-\sqrt{7}}{\sqrt{3}+\sqrt{5}-\sqrt{7}}=
\frac{\sqrt{3}+\sqrt{5}-\sqrt{7}}{(\sqrt{3}+\sqrt{5})^2-(\sqrt{7})^2}=\frac{\sqrt{3}+\sqrt{5}-\sqrt{7}}{3+2\sqrt{3}\sqrt{5}+5-7}=
\frac{\sqrt{3}+\sqrt{5}-\sqrt{7}}{1+2\sqrt{15}}=\frac{\sqrt{3}+\sqrt{5}-\sqrt{7}}{2\sqrt{15}+1}\times \frac{2\sqrt{15}-1}{2\sqrt{15}-1}=
\frac{(\sqrt{3}+\sqrt{5}-\sqrt{7})\times (2\sqrt{15}-1)}{(2\sqrt{15})^2-1^2}=\frac{2\sqrt{15}\times \sqrt{3}+2\sqrt{15}\times \sqrt{5}-2\sqrt{15}\times \sqrt{7}-1(\sqrt{3}+\sqrt{5}-\sqrt{7}}{60-1}=
\frac{2\times 3\sqrt{5}+2\times 5\sqrt{3}-2\sqrt{105}-\sqrt{3}-\sqrt{5}+\sqrt{7}}{59}=\frac{5\sqrt{5}+9\sqrt{3}+\sqrt{7}-2\sqrt{105}}{59}=
\frac{5\sqrt{5}+9\sqrt{3}+\sqrt{7}(1-2\sqrt{15})}{59}\)
Espero ter ajudado!
\(\frac{1}{\sqrt{3}+\sqrt{5}+\sqrt{7}}=\frac{1}{\sqrt{3}+\sqrt{5}+\sqrt{7}}\times \frac{\sqrt{3}+\sqrt{5}-\sqrt{7}}{\sqrt{3}+\sqrt{5}-\sqrt{7}}=
\frac{\sqrt{3}+\sqrt{5}-\sqrt{7}}{(\sqrt{3}+\sqrt{5})^2-(\sqrt{7})^2}=\frac{\sqrt{3}+\sqrt{5}-\sqrt{7}}{3+2\sqrt{3}\sqrt{5}+5-7}=
\frac{\sqrt{3}+\sqrt{5}-\sqrt{7}}{1+2\sqrt{15}}=\frac{\sqrt{3}+\sqrt{5}-\sqrt{7}}{2\sqrt{15}+1}\times \frac{2\sqrt{15}-1}{2\sqrt{15}-1}=
\frac{(\sqrt{3}+\sqrt{5}-\sqrt{7})\times (2\sqrt{15}-1)}{(2\sqrt{15})^2-1^2}=\frac{2\sqrt{15}\times \sqrt{3}+2\sqrt{15}\times \sqrt{5}-2\sqrt{15}\times \sqrt{7}-1(\sqrt{3}+\sqrt{5}-\sqrt{7}}{60-1}=
\frac{2\times 3\sqrt{5}+2\times 5\sqrt{3}-2\sqrt{105}-\sqrt{3}-\sqrt{5}+\sqrt{7}}{59}=\frac{5\sqrt{5}+9\sqrt{3}+\sqrt{7}-2\sqrt{105}}{59}=
\frac{5\sqrt{5}+9\sqrt{3}+\sqrt{7}(1-2\sqrt{15})}{59}\)
Espero ter ajudado!