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If

find \(\lfloor z \rfloor\).

 Aug 9, 2019
 #1
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\(\text{I'm assuming the braces stand for the fractional part of a number. $\{x\}=x-\lfloor x \rfloor$}\\ \{3\} = \sqrt{3}-1\\ \{2\}=\sqrt{2}-1\\~\\ z = \dfrac{(\sqrt{3}-1)^2-2(\sqrt{2}-1)^2}{(\sqrt{3}-1)-2(\sqrt{2}-1)}\)

 

\(z = \dfrac{(3-2\sqrt{3}+1)-2(2-2\sqrt{2}+1)}{\sqrt{3}-2\sqrt{2}+1}=\\ \dfrac{-2-2\sqrt{3}+4\sqrt{2}}{\sqrt{3}-2\sqrt{2}+1} = \\ -2\cdot \dfrac{\sqrt{3}-2\sqrt{2}+1}{\sqrt{3}-2\sqrt{2}+1} = -2\)

 

\(\lfloor -2 \rfloor = -2\)

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 Aug 9, 2019
edited by Rom  Aug 9, 2019

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