Odpowiedź:
[tex]c)\\a_1=\sqrt{12} =2\sqrt{3} \\\displaystyle q=\frac{-\sqrt{6} }{\sqrt{12} } =-\sqrt{\frac{6}{12} } =-\sqrt{\frac{1}{2} } =-\frac{1}{\sqrt{2} } =-\frac{\sqrt{2} }{2} \\S=\frac{a_1}{1-q} \\S=\frac{2\sqrt{3} }{1+\frac{\sqrt{2} }{2} } =\frac{2\sqrt{3} }{\frac{2+\sqrt{2} }{2} } =\frac{4\sqrt{3} }{2+\sqrt{2} } \cdot\frac{2-\sqrt{2} }{2-\sqrt{2} } =\frac{4\sqrt{3} (2-\sqrt{2}) }{4-2} =2\sqrt{3} (2-\sqrt{2)} =\\=\underline{4\sqrt{3}-2\sqrt{6} }[/tex]
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Odpowiedź:
[tex]c)\\a_1=\sqrt{12} =2\sqrt{3} \\\displaystyle q=\frac{-\sqrt{6} }{\sqrt{12} } =-\sqrt{\frac{6}{12} } =-\sqrt{\frac{1}{2} } =-\frac{1}{\sqrt{2} } =-\frac{\sqrt{2} }{2} \\S=\frac{a_1}{1-q} \\S=\frac{2\sqrt{3} }{1+\frac{\sqrt{2} }{2} } =\frac{2\sqrt{3} }{\frac{2+\sqrt{2} }{2} } =\frac{4\sqrt{3} }{2+\sqrt{2} } \cdot\frac{2-\sqrt{2} }{2-\sqrt{2} } =\frac{4\sqrt{3} (2-\sqrt{2}) }{4-2} =2\sqrt{3} (2-\sqrt{2)} =\\=\underline{4\sqrt{3}-2\sqrt{6} }[/tex]