Question

Uprość wyrażenie





Proszę o pełne rozwiązanie ze wszystkimi przekształceniami. Zachęcam do zobaczenia moich innych pytań z matematyki.

Answer 1

Verified answer

Odpowiedź:

[tex]\sqrt{2}[/tex]

Szczegółowe wyjaśnienie:

[tex]$\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}} +\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}[/tex]

Najpierw zauważmy, że:

[tex]$\sqrt{2+\sqrt{3}}=\frac{\sqrt{3}+1}{\sqrt{2}}[/tex]

[tex]$\sqrt{2-\sqrt{3}}=\frac{\sqrt{3}-1}{\sqrt{2}}[/tex]

Stąd:

[tex]$\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}} =\frac{2+\sqrt{3}}{\sqrt{2}+\frac{\sqrt{3}+1}{\sqrt{2}} }=\frac{2+\sqrt{3}}{\frac{2+\sqrt{3}+1}{\sqrt{2}} } =\frac{\sqrt{2}(2+\sqrt{3})}{3+\sqrt{3}} =[/tex]

[tex]$=\frac{\sqrt{2}(2+\sqrt{3})(3-\sqrt{3})}{9-3} =\frac{\sqrt{2}(6-2\sqrt{3}+3\sqrt{3}-3)}{6} =\frac{\sqrt{2}(3+\sqrt{3})}{6}=\frac{3\sqrt{2}+\sqrt{6}}{6}[/tex]

[tex]$\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\frac{2-\sqrt{3}}{\sqrt{2}-\frac{\sqrt{3}-1}{\sqrt{2}}}=\frac{2-\sqrt{3}}{\frac{2-\sqrt{3}+1}{\sqrt{2}} } =\frac{\sqrt{2}(2-\sqrt{3})}{3-\sqrt{3}} =[/tex]

[tex]$=\frac{\sqrt{2}(2-\sqrt{3})(3+\sqrt{3})}{9-3}=\frac{\sqrt{2}(6+2\sqrt{3}-3\sqrt{3}-3)}{6} =\frac{\sqrt{2}(3-\sqrt{3})}{6} =\frac{3\sqrt{2}-\sqrt{6}}{6}[/tex]

Ostatecznie:

[tex]$\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}} +\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\frac{3\sqrt{2}+\sqrt{6}}{6}+\frac{3\sqrt{2}-\sqrt{6}}{6}=\frac{6\sqrt{2}}{6}=\sqrt{2}[/tex]