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A= 1+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)
= \(\dfrac{2015}{2015}\)+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)
= 2015.(\(\dfrac{1}{2015}\)+\(\dfrac{1}{2014}\)+\(\dfrac{1}{2013}\)+...+\(\dfrac{1}{2}\))=2015.B
\(\Rightarrow\) \(\dfrac{A}{B}\)=2015
a) Ta có: \(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{2014}\right)\left(1-\dfrac{1}{2015}\right)\left(1-\dfrac{1}{2016}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2013}{2014}\cdot\dfrac{2014}{2015}\cdot\dfrac{2015}{2016}\)
\(=\dfrac{1}{2016}\)
b) Ta có: \(\dfrac{x-2}{12}+\dfrac{x-2}{20}+\dfrac{x-2}{30}+\dfrac{x-2}{42}+\dfrac{x-2}{56}+\dfrac{x-2}{72}=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right)=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{3}-\dfrac{1}{9}\right)=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\cdot\dfrac{2}{9}=\dfrac{16}{9}\)
\(\Leftrightarrow x-2=\dfrac{16}{9}:\dfrac{2}{9}=\dfrac{16}{9}\cdot\dfrac{9}{2}=8\)
hay x=10
Vậy: x=10
Lời giải:
Ta có:
\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}\)
\(S> \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2015.2016}\)
\(\Leftrightarrow S> \frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{2016-2015}{2015.2016}\)
\(\Leftrightarrow S> \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(\Leftrightarrow S> \frac{1}{2}-\frac{1}{2016}=\frac{1007}{2016}\)
--------------------------
\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{2015^2}\)
\(S< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{2014}{2015}\)
\(\Leftrightarrow S< \frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{2015-2014}{2014.2015}\)
\(\Leftrightarrow S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{2014}-\frac{1}{2015}\)
\(\Leftrightarrow S< 1-\frac{1}{2015}=\frac{2014}{2015}\)
Vậy ta có đpcm.
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!!! ~
1,
đặt A= \(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+....+\(\dfrac{1}{2016}\)+\(\dfrac{1}{2017}\)
2A=1+\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+....+\(\dfrac{1}{2015}\)+\(\dfrac{1}{2016}\)
2A-A=(1+\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+....+\(\dfrac{1}{2015}\)+\(\dfrac{1}{2016}\))-(\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+....+\(\dfrac{1}{2016}\)+\(\dfrac{1}{2017}\))
A=1-\(\dfrac{1}{2017}\)
A=\(\dfrac{2016}{2017}\)
vậy A=\(\dfrac{2016}{2017}\)
Đặt \(S=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2016}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{1008}\right)\)
\(=\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\)
Nên:
\(A=\left(\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right):\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\right)\)\(=\left(\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right):\left(\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right)\)\(\Rightarrow A=1\)
Vậy A = 1
Chúc bạn học tốt!!
\(A=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2015}}\)
\(\Rightarrow5A=1+\dfrac{1}{5}+...+\dfrac{1}{5^{2014}}\)
\(\Rightarrow5A-A=\left(1+\dfrac{1}{5}+...+\dfrac{1}{5^{2014}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2015}}\right)\)
\(\Rightarrow4A=1-\dfrac{1}{5^{2015}}\)
\(\Rightarrow A=\dfrac{1}{4}-\dfrac{1}{5^{2015}.4}< \dfrac{1}{4}\)
\(\Rightarrowđpcm\)
\(A=\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2014}}+\dfrac{1}{5^{2015}}\)
\(\Rightarrow5A=5\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2014}}+\dfrac{1}{5^{2015}}\right)\)
\(\Rightarrow5A=1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2013}}+\dfrac{1}{5^{2014}}\)
\(\Rightarrow5A-A=\left(1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2013}}+\dfrac{1}{5^{2014}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2014}}+\dfrac{1}{5^{2015}}\right)\)
\(\Rightarrow4A=1-\dfrac{1}{5^{2015}}\)
\(\Rightarrow A=\dfrac{1}{4}-\dfrac{1}{5^{2015}.4}\)
\(\Rightarrow A< \dfrac{1}{4}\)
Ta có :
B = \(\dfrac{2015}{1}+\dfrac{2014}{2}+\dfrac{2013}{3}+...+\dfrac{2}{2014}+\dfrac{1}{2015}\) => B = \(\left(1+\dfrac{2014}{2}\right)+\left(1+\dfrac{2013}{3}\right)+...+\left(1+\dfrac{2}{2014}\right)+\left(1+\dfrac{1}{2015}\right)+1\) => B = \(\dfrac{2016}{2}+\dfrac{2016}{3}+...+\dfrac{2016}{2014}+\dfrac{2016}{2015}+\dfrac{2016}{2016}\) => B = \(2016\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right)\) Ta có :
\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}}{2016\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}\right)}\)
=> \(\dfrac{A}{B}=\dfrac{1}{2016}\)
Vậy \(\dfrac{A}{B}=\dfrac{1}{2016}\)
cảm ơn bạn nhiều nhé