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\(A=\dfrac{\dfrac{1}{2017}+\dfrac{2}{2016}+\dfrac{3}{2015}+...+\dfrac{2016}{2}+\dfrac{2017}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{\left(\dfrac{1}{2017}+1\right)+\left(\dfrac{2}{2016}+1\right)+\left(\dfrac{3}{2015}+1\right)+...+\left(\dfrac{2016}{2}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{\dfrac{2018}{2017}+\dfrac{2018}{2016}+\dfrac{2018}{2015}+...+\dfrac{2018}{2}+\dfrac{2018}{2018}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{2018\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}=2018\)
Bài 2:
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2016}{2017}\)
\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2016}{2017}\)
\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{2016}{2017}\)
\(\Leftrightarrow\dfrac{1}{x+1}=1-\dfrac{2016}{2017}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2017}\)
\(\Leftrightarrow x+1=2017\Leftrightarrow x=2016\)
Vậy \(x=2016\)
Ta thấy A > 0 (1)
Vì \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{2016^2}< \dfrac{1}{2015.2016}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2015.2016}\)
\(\Rightarrow A>1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2015}-\dfrac{1}{2016}=1-\dfrac{1}{2016}=\dfrac{2015}{2016}< 1\)(2)
Từ (1)(2) => 0 < A < 1
Vậy A không phải là số tự nhiên
Giải:
Ta có: \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2016^2}>0_{\left(1\right)}.\) (do A là phân số dương).
Ta lại có:
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2016^2}.\)
\(=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{2016.2016}.\)
\(< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2015.2016}.\)
\(< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}.\)
\(< 1+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{2015}-\dfrac{1}{2015}\right)-\dfrac{1}{2016}.\)\(< 1+0+0+0+...+0-\dfrac{1}{2016}.\)
\(< 1-\dfrac{1}{2016}.\)
\(< \dfrac{2015}{2016}.\)
\(\Rightarrow A< 1_{\left(2\right)}.\) (do \(\dfrac{2015}{2016}< 1\)).
Từ \(_{\left(1\right)}\) và \(_{\left(2\right)}\) \(\Rightarrow0< A< 1.\)
\(\Rightarrow A\) không phải là số tự nhiên.
Vậy ta thu được \(đpcm.\)
~ Học tốt!!! ~
A < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{2016.2017}\)
=> A<\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2016}-\dfrac{1}{2017}\)
=> A<\(1-\dfrac{1}{2017}\)
Vì \(\dfrac{1}{2017}>\dfrac{1}{2017^2.2018^2}\) nên \(1-\dfrac{1}{2017}< 1-\dfrac{1}{2017^2.2018^2}\)
=> A<\(\dfrac{1}{2017}\)<B
Vậy A < B
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