Question

$5-2 \sqrt{6} = \frac{a \sqrt{2}+b \sqrt{3} }{ \sqrt{2} + \sqrt{3} }$ aflati valoarea raportului  $\frac{a}{b}$

Answer 1

$5-2 \sqrt{6} = \frac{a \sqrt{2}+b \sqrt{3} }{ \sqrt{2}+ \sqrt{3} } \Rightarrow (5- 2\sqrt{6} )( \sqrt{2}+ \sqrt{3} )=a \sqrt{2}+b \sqrt{3} \Leftrightarrow \\ \Leftrightarrow 5 \sqrt{2}+5\sqrt{3}-4 \sqrt{3}-6 \sqrt{2} =a \sqrt{2}+b \sqrt{3} \Leftrightarrow \\ \Leftrightarrow- \sqrt{2}+ \sqrt{3}=a \sqrt{2}+b \sqrt{3} \Leftrightarrow \\ \Leftrightarrow \sqrt{3}-b \sqrt{3}=a \sqrt{2}+ \sqrt{2} \Leftrightarrow \\ \Leftrightarrow \sqrt{3}(1-b)= \sqrt{2}(a+1)$

$\sqrt{3}(1-b)= \sqrt{2}(a+1) \Rightarrow \frac{ \sqrt{3} (1-b)}{ \sqrt{ 2}}=a+1,~dar~(a+1)~si~(b-1) \\ sunt~\underline{rationale},~iar~ \frac{ \sqrt{3} }{ \sqrt{2} }~este~\underline{irational };~deci~pentru~ca~egalitatea \\ sa~aiba~loc,~trebuie~ca~a+1=b-1=0 \Rightarrow a=-1~si~b=1. \\ \\ \frac{a}{b}= \frac{-1}{1}=-1.$

Answer 2

Folosim proprietatea fundamentala a proportilor(produsul mezilor este egal cu produsul extremilor):
$a\sqrt{2}+b\sqrt{3}=(5-2\sqrt{6})(\sqrt{2}+\sqrt{3}) = 5\sqrt{2} + 5\sqrt{3} -4\sqrt{3} - 6\sqrt{2}=\\=\ \textgreater \ a\sqrt{2}+b{\sqrt{3}=-\sqrt{2}+\sqrt{3} =\ \textgreater \ a=-1, \\\\b=1$
Atunci, a/b=-1/1=-1

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