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	16	thêm 11 ký tự trong đáp án		Sửa 07-06-14 07:22 AM vuvanbang70 1
ta có $\frac{1}{\sqrt{k}}$ = $\frac{2}{\sqrt{k} + \sqrt{k}}$ > $\frac{2}{\sqrt{k+1}+\sqrt{k}}$ = $2(\sqrt{k+1}-\sqrt{k})\Rightarrow\frac{1}{\sqrt{2}}...$>17tương tự $\frac{1}{\sqrt{k}}$ < $2(\sqrt{k}-\sqrt{k-1}) \Rightarrow\frac{1}{\sqrt{2}}+...$<18 ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}} =2(\sqrt{k+1}-\sqrt{k})\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{\sqrt{k}} < 2(\sqrt{k}-\sqrt{k-1}) \Rightarrow\frac{1}{\sqrt{2}}+...<18$ ta có $\frac{1}{\sqrt{k}}$ = $\frac{2}{\sqrt{k} + \sqrt{k}}$ > $\frac{2}{\sqrt{k+1}+\sqrt{k}}$ = $2(\sqrt{k+1}-\sqrt{k})\Rightarrow\frac{1}{\sqrt{2}}...$>17tương tự $\frac{1}{\sqrt{k}}$ < $2(\sqrt{k}-\sqrt{k-1}) \Rightarrow\frac{1}{\sqrt{2}}+...$<18

	15	sửa đáp án		Sửa 07-06-14 07:18 AM vuvanbang70 1
ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}} =2(\sqrt{k+1}-\sqrt{k})\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{\sqrt{k}} < 2(\sqrt{k}-\sqrt{k-1}) \Rightarrow\frac{1}{\sqrt{2}}+...<18$ ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}} =2(\sqrt{k+1}-\sqrt{k})\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k} < 2(\sqrt{\sqrt{k}}-\sqrt{k-1}) \Rightarrow\frac{1}{\sqrt{2}}+...<18$ ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}} =2(\sqrt{k+1}-\sqrt{k})\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{\sqrt{k}} < 2(\sqrt{k}-\sqrt{k-1}) \Rightarrow\frac{1}{\sqrt{2}}+...<18$

	14	thêm 7 ký tự trong đáp án		Sửa 07-06-14 07:16 AM vuvanbang70 1
ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}} =2(\sqrt{k+1}-\sqrt{k})\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k} < 2(\sqrt{\sqrt{k}}-\sqrt{k-1}) \Rightarrow\frac{1}{\sqrt{2}}+...<18$ ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}} =2(\sqrt{k+1}-\sqrt{k})\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k} < 2(\sqrt{k}-\sqrt{k-1}) \Rightarrow\frac{1}{\sqrt{2}}+...<18$ ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}} =2(\sqrt{k+1}-\sqrt{k})\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k} < 2(\sqrt{\sqrt{k}}-\sqrt{k-1}) \Rightarrow\frac{1}{\sqrt{2}}+...<18$

	13	bớt 2 ký tự trong đáp án		Sửa 07-06-14 07:10 AM vuvanbang70 1
ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}} =2(\sqrt{k+1}-\sqrt{k})\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k} < 2(\sqrt{k}-\sqrt{k-1}) \Rightarrow\frac{1}{\sqrt{2}}+...<18$ ta có $ \frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}} =2(\sqrt{k+1}-\sqrt{k})\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k} < 2(\sqrt{k}-\sqrt{k-1}) \Rightarrow\frac{1}{\sqrt{2}}+...<18 $ ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}} =2(\sqrt{k+1}-\sqrt{k})\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k} < 2(\sqrt{k}-\sqrt{k-1}) \Rightarrow\frac{1}{\sqrt{2}}+...<18$

	12	thêm 2 ký tự trong đáp án		Sửa 07-06-14 07:10 AM vuvanbang70 1
ta có $ \frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}} =2(\sqrt{k+1}-\sqrt{k})\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k} < 2(\sqrt{k}-\sqrt{k-1}) \Rightarrow\frac{1}{\sqrt{2}}+...<18 $ ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}} =2(\sqrt{k+1}-\sqrt{k})\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k} < 2(\sqrt{k}-\sqrt{k-1}) \Rightarrow\frac{1}{\sqrt{2}}+...<18$ ta có $ \frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}} =2(\sqrt{k+1}-\sqrt{k})\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k} < 2(\sqrt{k}-\sqrt{k-1}) \Rightarrow\frac{1}{\sqrt{2}}+...<18 $

	11	thêm 1 ký tự trong đáp án		Sửa 05-06-14 12:09 PM vuvanbang70 1
ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}} =2(\sqrt{k+1}-\sqrt{k})\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k} < 2(\sqrt{k}-\sqrt{k-1}) \Rightarrow\frac{1}{\sqrt{2}}+...<18$ ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}} =2(\sqrt{k+1}-\sqrt{k})\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k} < 2(\sqrt{k}-\sqrt{k-1}) \Rightarrow\frac{1}{\sqrt{2}}...<18$ ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}} =2(\sqrt{k+1}-\sqrt{k})\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k} < 2(\sqrt{k}-\sqrt{k-1}) \Rightarrow\frac{1}{\sqrt{2}}+...<18$

	10	sửa đáp án		Sửa 05-06-14 12:08 PM vuvanbang70 1
ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}} =2(\sqrt{k+1}-\sqrt{k})\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k} < 2(\sqrt{k}-\sqrt{k-1}) \Rightarrow\frac{1}{\sqrt{2}}...<18$ ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}} =2(\sqrt{k+1}-\sqrt{k}\Rightarrow\frac{1}{\sqrt{2}})...>17tương tự \frac{1}{k} < 2(\sqrt{k}-\sqrt{k-1}) \Rightarrow\frac{1}{\sqrt{2}}...<18$ ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}} =2(\sqrt{k+1}-\sqrt{k})\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k} < 2(\sqrt{k}-\sqrt{k-1}) \Rightarrow\frac{1}{\sqrt{2}}...<18$

	9	sửa đáp án		Sửa 05-06-14 12:07 PM vuvanbang70 1
ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}} =2(\sqrt{k+1}-\sqrt{k}\Rightarrow\frac{1}{\sqrt{2}})...>17tương tự \frac{1}{k} < 2(\sqrt{k}-\sqrt{k-1}) \Rightarrow\frac{1}{\sqrt{2}}...<18$ ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}} =2(\sqrt{k+1}-\sqrt{k}\Rightarrow\frac{1}{\sqrt{2}})...>17tương tự \frac{1}{k} < 2(\sqrt{k}-\sqrt{k-1}) \Rightarrow\frac{1}{\sqrt{2}}...<18$ ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}} =2(\sqrt{k+1}-\sqrt{k}\Rightarrow\frac{1}{\sqrt{2}})...>17tương tự \frac{1}{k} < 2(\sqrt{k}-\sqrt{k-1}) \Rightarrow\frac{1}{\sqrt{2}}...<18$

	8	thêm 1 ký tự trong đáp án		Sửa 05-06-14 12:06 PM vuvanbang70 1
ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}} =2(\sqrt{k+1}-\sqrt{k}\Rightarrow\frac{1}{\sqrt{2}})...>17tương tự \frac{1}{k} < 2(\sqrt{k}-\sqrt{k-1}) \Rightarrow\frac{1}{\sqrt{2}}...<18$ ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}} =2(\sqrt{k+1}-\sqrt{k}\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k} < 2(\sqrt{k}-\sqrt{k-1}) \Rightarrow\frac{1}{\sqrt{2}}...<18$ ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}} =2(\sqrt{k+1}-\sqrt{k}\Rightarrow\frac{1}{\sqrt{2}})...>17tương tự \frac{1}{k} < 2(\sqrt{k}-\sqrt{k-1}) \Rightarrow\frac{1}{\sqrt{2}}...<18$

	7	thêm 1 ký tự trong đáp án		Sửa 05-06-14 12:05 PM vuvanbang70 1
ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}} =2(\sqrt{k+1}-\sqrt{k}\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k} < 2(\sqrt{k}-\sqrt{k-1}) \Rightarrow\frac{1}{\sqrt{2}}...<18$ ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}} =2(\sqrt{k+1}-\sqrt{k}\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k} < 2(\sqrt{k}-\sqrt{k-1} \Rightarrow\frac{1}{\sqrt{2}}...<18$ ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}} =2(\sqrt{k+1}-\sqrt{k}\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k} < 2(\sqrt{k}-\sqrt{k-1}) \Rightarrow\frac{1}{\sqrt{2}}...<18$

	6	thêm 1 ký tự trong đáp án		Sửa 05-06-14 12:05 PM vuvanbang70 1
ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}} =2(\sqrt{k+1}-\sqrt{k}\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k} < 2(\sqrt{k}-\sqrt{k-1} \Rightarrow\frac{1}{\sqrt{2}}...<18$ ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}}=2(\sqrt{k+1}-\sqrt{k}\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k} < 2(\sqrt{k}-\sqrt{k-1} \Rightarrow\frac{1}{\sqrt{2}}...<18$ ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}} =2(\sqrt{k+1}-\sqrt{k}\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k} < 2(\sqrt{k}-\sqrt{k-1} \Rightarrow\frac{1}{\sqrt{2}}...<18$

	5	bớt 1 ký tự trong đáp án		Sửa 05-06-14 12:04 PM vuvanbang70 1
ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}}=2(\sqrt{k+1}-\sqrt{k}\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k} < 2(\sqrt{k}-\sqrt{k-1} \Rightarrow\frac{1}{\sqrt{2}}...<18$ ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}}=2(\sqrt{k+1}-\sqrt{k}\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k} < \2(\sqrt{k}-\sqrt{k-1} \Rightarrow\frac{1}{\sqrt{2}}...<18$ ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}}=2(\sqrt{k+1}-\sqrt{k}\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k} < 2(\sqrt{k}-\sqrt{k-1} \Rightarrow\frac{1}{\sqrt{2}}...<18$

	4	thêm 3 ký tự trong đáp án		Sửa 05-06-14 12:04 PM vuvanbang70 1
ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}}=2(\sqrt{k+1}-\sqrt{k}\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k} < \2(\sqrt{k}-\sqrt{k-1} \Rightarrow\frac{1}{\sqrt{2}}...<18$ ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}}=2(\sqrt{k+1}-\sqrt{k}\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k}\leq2(\sqrt{k}-\sqrt{k-1} \Rightarrow\frac{1}{\sqrt{2}}...<18$ ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} > \frac{2}{\sqrt{k+1}+\sqrt{k}}=2(\sqrt{k+1}-\sqrt{k}\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k} < \2(\sqrt{k}-\sqrt{k-1} \Rightarrow\frac{1}{\sqrt{2}}...<18$

	3	sửa đáp án		Sửa 05-06-14 12:02 PM vuvanbang70 1
ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} &gt; \frac{2}{\sqrt{k+1}+\sqrt{k}}=2(\sqrt{k+1}-\sqrt{k}\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k}\leq2(\sqrt{k}-\sqrt{k-1} \Rightarrow\frac{1}{\sqrt{2}}...<18$ ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} \geq \frac{2}{\sqrt{k+1}+\sqrt{k}}=2(\sqrt{k+1}-\sqrt{k}\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k}\leq2(\sqrt{k}-\sqrt{k-1} \Rightarrow\frac{1}{\sqrt{2}}...<18$ ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} &gt; \frac{2}{\sqrt{k+1}+\sqrt{k}}=2(\sqrt{k+1}-\sqrt{k}\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k}\leq2(\sqrt{k}-\sqrt{k-1} \Rightarrow\frac{1}{\sqrt{2}}...<18$

	2	thêm 2 ký tự trong đáp án		Sửa 05-06-14 12:01 PM vuvanbang70 1
ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} \geq \frac{2}{\sqrt{k+1}+\sqrt{k}}=2(\sqrt{k+1}-\sqrt{k}\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k}\leq2(\sqrt{k}-\sqrt{k-1} \Rightarrow\frac{1}{\sqrt{2}}...<18$ ta có \frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} \geq \frac{2}{\sqrt{k+1}+\sqrt{k}}=2(\sqrt{k+1}-\sqrt{k}\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k}\leq2(\sqrt{k}-\sqrt{k-1} \Rightarrow\frac{1}{\sqrt{2}}...<18 ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} \geq \frac{2}{\sqrt{k+1}+\sqrt{k}}=2(\sqrt{k+1}-\sqrt{k}\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k}\leq2(\sqrt{k}-\sqrt{k-1} \Rightarrow\frac{1}{\sqrt{2}}...<18$

	1	tạo mới		Trả lời 05-06-14 12:00 PM vuvanbang70 1
ta có \frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} \geq \frac{2}{\sqrt{k+1}+\sqrt{k}}=2(\sqrt{k+1}-\sqrt{k}\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k}\leq2(\sqrt{k}-\sqrt{k-1} \Rightarrow\frac{1}{\sqrt{2}}...<18