Răspuns:
vom folosi proprietatea (a - b)² = (b - a)²
care se demonstreaza astfel:
(a - b)² =(a - b) · (a - b) = (-1) · (b - a) · (-1) · (b - a) = 1 · (b - a) · (b - a) = (b - a)²
si formula (a + b) · (a - b) = a² - b²
[tex]\frac{\sqrt{5} -\sqrt{3} }{\sqrt{5} +\sqrt{3}} :(\frac{2}{\sqrt{3} -\sqrt{5}} )^{-2} =\frac{\sqrt{5} -\sqrt{3} }{\sqrt{5} +\sqrt{3}} :(\frac{\sqrt{3} -\sqrt{5}}{2} )^{2}=\frac{\sqrt{5} -\sqrt{3} }{\sqrt{5} +\sqrt{3}} *(\frac{2}{\sqrt{3} -\sqrt{5}} )^{2}=[/tex]
[tex]=\frac{\sqrt{5} -\sqrt{3} }{\sqrt{5} +\sqrt{3}} *\frac{2^{2}}{(\sqrt{3} -\sqrt{5})^{2} } =\frac{\sqrt{5} -\sqrt{3} }{\sqrt{5} +\sqrt{3}} *\frac{4}{(\sqrt{5} -\sqrt{3})^{2} } =[/tex]
[tex]=\frac{\sqrt{5} -\sqrt{3} }{\sqrt{5} +\sqrt{3}} *\frac{4}{(\sqrt{5} -\sqrt{3})(\sqrt{5} -\sqrt{3}) } =\frac{1 }{\sqrt{5} +\sqrt{3}} *\frac{4}{(\sqrt{5} -\sqrt{3}) } =[/tex]
[tex]=\frac{4}{(\sqrt{5} +\sqrt{3})(\sqrt{5} -\sqrt{3}) }=\frac{4}{(\sqrt{5})^2 -(\sqrt{3})^2}=\frac{4}{5 -3}=\frac{4}{2} =2[/tex]
2 ∈ (√3, √5)
Explicație pas cu pas:
