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A = \(\frac{2015.2016-1}{2015.2016}\)= \(\frac{2015.2016}{2015.2016}\)\(-\)\(\frac{1}{2015.2016}\)= 1 \(-\)\(\frac{1}{2015.2016}\)
B = \(\frac{2016.2017-1}{2016.2017}\)= \(\frac{2016.2017}{2016.2017}\)\(-\)\(\frac{1}{2016.2017}\)= 1 \(-\)\(\frac{1}{2016.2017}\)
Vì \(\frac{1}{2015.2016}\)> \(\frac{1}{2016.2017}\)
=> 1 \(-\)\(\frac{1}{2015.2016}\)< \(1-\)\(\frac{1}{2016.2017}\)
=> A < B
A=\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{2014\cdot2015\cdot2016}=\dfrac{1}{2}\cdot\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{2014\cdot2015}-\dfrac{1}{2015\cdot2016}\right)=\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{2015}\cdot\dfrac{1}{2016}\right)=\dfrac{1}{4}-\dfrac{1}{2\cdot2015\cdot2016}< \dfrac{1}{4}\)
Vậy A<\(\dfrac{1}{4}\)
---bé hơn hoặc bằng tức là chỉ cần xảy ra 1 khả năng cũng thõa mãn nhé---
Bạn tham khảo: Câu hỏi của chipchip - Toán lớp 6 - Học toán với OnlineMath
A\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2015}-\frac{1}{2016}\)
A=\(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2015}-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
A=\(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2015}+\frac{1}{2016}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
A=\(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2015}+\frac{1}{2016}-\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1008}\right)\)
A=\(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2015}+\frac{1}{2016}\)
B-A=\(\left(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)-\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2015}+\frac{1}{2016}\right)\)
B-A=1/1008
P\(=\dfrac{3}{\left(1.2\right)^2}+\dfrac{5}{\left(2.3\right)^2}+.....+\dfrac{4033}{\left(2016.2017\right)^2}\) \(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+.......+\dfrac{4033}{2016^2.2017^2}\) \(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+....+\dfrac{1}{2016^2}-\dfrac{1}{2017^2}\) =1\(-\dfrac{1}{2017^2}\) Do `1\(-\dfrac{1}{2017^2}\) <1\(\Rightarrow\) P<1 ( ĐPCM)
P = \(\dfrac{3}{\left(1.2\right)^2}+\dfrac{5}{\left(2.3\right)^2}+\dfrac{7}{\left(3.4\right)^2}+...+\dfrac{4033}{\left(2016.2017\right)^2}\)
P = \(\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+...+\dfrac{4033}{\left(2016.2017\right)^2}\)
P = \(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{2016^2}-\dfrac{1}{2017^2}\)
P = \(1-\dfrac{1}{2017^2}\)
⇒ P < 1
⇒ ĐPCM
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