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Rozwiązane

Proszę o rozwiązanie.

Proszę O Rozwiązanie class=

Odpowiedź :

Magdago

Odpowiedź:

[tex]\dfrac{\sqrt{a}+\sqrt{b}}{a-b}:\dfrac{a\cdot b}{\sqrt{a}-\sqrt{b}}=\dfrac{\sqrt{a}+\sqrt{b}}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}\cdot\dfrac{\sqrt{a}-\sqrt{b}}{a\cdot b}=\\\\\\=\dfrac{1}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\sqrt{a}-\sqrt{b}}{a\cdot b}=\dfrac{1}{a\cdot b}\\\\\\\\dla\ \ a=3+\sqrt{2}\ \ \ \ i\ \ \ \ b=3-\sqrt{2}\\\\\\\dfrac{1}{a\cdot b}=\dfrac{1}{(3+\sqrt{2})(3-\sqrt{2})}=\dfrac{1}{3^2-(\sqrt{2})^2}=\dfrac{1}{9-2}=\dfrac{1}{7}\\\\\\Odp.A[/tex]

Mushriimgo

[tex]a=3+\sqrt{2} \\b = 3 - \sqrt{2} \\\frac{\sqrt{a} + \sqrt{b} }{a-b} : \frac{a*b}{\sqrt{a}-\sqrt{b} }=\frac{\sqrt{a} + \sqrt{b} }{a-b} *\frac{\sqrt{a}-\sqrt{b}}{a*b} = \frac{(\sqrt{a} + \sqrt{b})(\sqrt{a}-\sqrt{b})}{(a-b)ab} = \frac{(\sqrt{a})^{2} - (\sqrt{b} )^{2} }{(a-b)ab} = \frac{|a| - |b|}{(a-b)ab}[/tex]

[tex]\frac{|3+\sqrt{2}| - |3-\sqrt{2}| }{[3+\sqrt{2}-(3-\sqrt{2})]*(3+\sqrt{2})(3-\sqrt{2})} = \frac{3+\sqrt{2}-(3-\sqrt{2})}{2\sqrt{2}(9-2)} = \frac{2\sqrt{2}}{14\sqrt{2}} = \frac{1}{7}[/tex]

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