TÍNH : \(\frac{1\cdot2016+2\cdot2015+...+2015\cdot2+2016\cdot1}{1\cdot2+2\cdot3+...+2016\cdot2017}\)
Những câu hỏi liên quan
Tìm K biết :
K - 2016 = \(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2017\right)}{2017\cdot1+2016\cdot2+2015\cdot3+...+2\cdot2016+2\cdot2017}\)
Tính
\(1\cdot2\cdot3+2\cdot3\cdot4+3\cdot4\cdot5+...+2015\cdot2016\cdot2017\)
Đặt \(A=1.2.3+2.3.4+3.4.5+...+2015.2016.2017\)
=>\(4A=1.2.3.4+2.3.4.4+3.4.5.4+...+2015.2016.2017.4\)
=>\(4A=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+3.4.5.\left(6-2\right)\)
\(+...+2015.2016.2017.\left(2018-2014\right)\)
=>\(4A=1.2.3.4-0.1.2.3+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5\)
\(+...+2015.2016.2017.2018-2014.2015.2016.2017\)
=>\(4A=2015.2016.2017.2018\Rightarrow A=\frac{2015.2016.2017.2018}{4}\)
Đúng 0
Bình luận (0)
Tính giá trị biểu thức: B=\(\frac{x}{1\cdot2}+\frac{x}{2\cdot3}+\frac{x}{3\cdot4}+...+\frac{x}{2015\cdot2016}\) tại x=2016
Thay x = 2016 vào biểu thức B ta có :
\(B=\frac{2016}{1.2}+\frac{2016}{2.3}+\frac{2016}{3.4}+...+\frac{2016}{2015.2016}\)
\(B=2016\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2016}\right)\)
\(B=2016\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\right)\)
\(B=2016\left(1-\frac{1}{2016}\right)\)
\(B=2016.\frac{2015}{2016}\)
\(B=2015\)
Đúng 0
Bình luận (0)
Tính
\(A=1\cdot2\cdot3+2\cdot3\cdot4+3\cdot4\cdot5+...+2014\cdot2015\cdot2016\)
\(A=1.2.3+2.3.4+3.4.5+....+2014.2015.2016\)
\(4A=1.2.3.4+2.3.4.\left(5-1\right)+3.4.5.\left(6-2\right)+....+2014.2015.2016.\left(2017-2013\right)\)\(4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...+2014.2015.2016.2017-2013.2014.2015.2016\)\(4A=2014.2015.2016.2017\)
\(A=\dfrac{2014.2015.2016.2017}{4}=4215446423280\)
Đúng 0
Bình luận (0)
\(\frac{45\cdot16-17}{45\cdot15+28}\) các bạn đề là tính nhanh nhé
Ví dụ: \(\frac{2016\cdot2015-5}{2014\cdot2016+2011}=\frac{2016\cdot\left(2014+1\right)-5}{2014\cdot2016+2011}=\frac{2016\cdot2014+2016\cdot1-5}{2014\cdot2016-2011}=\frac{2016\cdot2014+2011}{2014\cdot2016+2011}=1\)
so sánh A và B
A= \(\frac{2015\cdot2016-1}{2015\cdot2016}\) B= \(\frac{2016\cdot2017-1}{2016\cdot2017}\)
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
Đúng 0
Bình luận (0)
Cho \(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{2014\cdot2015\cdot2016}\).
Chứng minh \(A\le\dfrac{1}{4}\).
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---
Đúng 0
Bình luận (0)
Cho \(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{2014\cdot2015\cdot2016}\).
So sánh A với \(\dfrac{1}{4}\).
tính:\(\frac{1\cdot98+2\cdot97+3\cdot96+...+97\cdot2+98\cdot1}{1\cdot2+2\cdot3+3\cdot4+...+99\cdot100}\)