Question

Simplify the radical expression

\dfrac{\sqrt{15}+\sqrt{35}+\sqrt{21}+5 }{\sqrt{3}+2\sqrt{5}+\sqrt{7} }

Collected in the board: Square Root

Steven Zheng posted 1 year ago


Answer

\dfrac{\sqrt{15}+\sqrt{35}+\sqrt{21}+5 }{\sqrt{3}+2\sqrt{5}+\sqrt{7} }

=\dfrac{\sqrt{3}\sqrt{5}+\sqrt{5}\sqrt{7}+\sqrt{3}\sqrt{7}+5}{\sqrt{3}+2\sqrt{5}+\sqrt{7} }

=\dfrac{\sqrt{5}(\sqrt{3}+\sqrt{5} ) +\sqrt{7}(\sqrt{3}+\sqrt{5} ) }{\sqrt{3}+2\sqrt{5}+\sqrt{7}}

=\dfrac{(\sqrt{3}+\sqrt{5} )(\sqrt{5}+\sqrt{7} ) }{\sqrt{3}+\sqrt{5}+\sqrt{5}+\sqrt{7}}

=\dfrac{1}{\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7} } }

=\dfrac{1}{\dfrac{1}{\sqrt{3}+\sqrt{5}}\cdotp \dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}+ \dfrac{1}{\sqrt{5}+\sqrt{7} }\cdotp \dfrac{\sqrt{7}-\sqrt{5} }{\sqrt{7}-\sqrt{5} } }

=\dfrac{1}{\dfrac{\sqrt{5}-\sqrt{3} }{2} +\dfrac{\sqrt{7}-\sqrt{5} }{2} }

=\dfrac{2}{\sqrt{7}-\sqrt{3} }

=\dfrac{\sqrt{7}+\sqrt{3} }{2}

Steven Zheng posted 2 weeks ago

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