Răspuns:
✿ Salut! ✿
✎ Cerință:
a) Arătați că [tex]a=\frac{1}{2}[/tex]
b) Arătați că numărul [tex]N=(b-2a)^{2}-\sqrt{24}\in \mathbb{N}[/tex]
✯✯✯ Rezolvare: ✯✯✯
a) Arătați că [tex]a=\frac{1}{2}[/tex].
[tex]a=\frac{\sqrt{2}-1}{\sqrt{2}}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{6}}+\frac{\sqrt{4}-\sqrt{3}}{\sqrt{12}}\\\\a=\frac{2-\sqrt{2}}{2}+\frac{\sqrt{18}-\sqrt{12}}{6}+\frac{2-\sqrt{3}}{2\sqrt{3}}\\\\a=\frac{6-3\sqrt{2}}{6}+\frac{3\sqrt{2}-2\sqrt{3}}{6}+\frac{2\sqrt{3}-3}{6}\\\\a=\frac{6-3\sqrt{2}+3\sqrt{2}-2\sqrt{3}+2\sqrt{3}-3}{6}\\\\a=\frac{\not3}{\not6}\\\\a=\frac{1}{2}[/tex]
b) Arătați că numărul [tex]N=(b-2a)^{2}-\sqrt{24}\in \mathbb{N}[/tex].
[tex]b = (\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{6}}):\frac{1}{\sqrt{6}}\\\\b=(\frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{3}+\frac{\sqrt{6}}{6})*\sqrt{6}\\\\b=(\frac{3\sqrt{2}}{6}+\frac{2\sqrt{3}}{6}+\frac{\sqrt{6}}{6})*\sqrt{6}\\\\b=\frac{3\sqrt{2}*\sqrt{6}}{6}+\frac{2\sqrt{3}*\sqrt{6}}{6}+\frac{6}{6}\\\\b=\frac{3\sqrt{12}}{6}+\frac{2\sqrt{18}}{6}+\frac{6}{6}\\\\b=\frac{\not3*\not2\sqrt{3}}{\not6}+\frac{\not2*\not3\sqrt{2}}{\not6}+1\\\\b=\sqrt{3}+\sqrt{2}+1[/tex]
[tex]N=(\sqrt{3}+\sqrt{2}+1-\not2*\frac{1}{\not2})^{2}-\sqrt{24}\\\\N=(\sqrt{3}+\sqrt{2}+1-1)^{2}-2\sqrt{6}\\\\N=(\sqrt{3}+\sqrt{2})^{2}-2\sqrt{6}\\\\N=(\sqrt{3})^{2}+2*\sqrt{3}*\sqrt{2}+(\sqrt{2})^{2}-2\sqrt{6}\\\\N=3+2\sqrt{6}+2-2\sqrt{6}\\\\N=5\in\mathbb{N}\\\\\implies N\in\mathbb{N}[/tex]
