Cho \(A=\dfrac{2016.2017-2}{2015+2015.2017},B=\dfrac{-2016.20172017}{20162016.2017}\)
Tinh A+B
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VN
13 tháng 5 2017
- Tính A trước:
\(A=\frac{\left(2015+1\right).2017-2}{2015+2015.2017}\)
\(A=\frac{2017.2015+2017-2}{2017.2015+2015}\)
\(A=\frac{2017.2015+2015}{2017.2015+2015}\)
\(A=1\)
- Tính B:
\(B=\frac{-2016.20172017}{2017.20162016}\)
\(B=\frac{-2016}{2017}.\frac{20172017}{20162016}\)
\(B=\frac{-10001}{10001}\)
\(B=-1\)
Vậy ta có: \(A+B=1-1=0\)
28 tháng 5 2017
a)
\(A=2\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{2015\cdot2017}\right)\)
\(=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2015\cdot2017}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\)
\(=1-\dfrac{1}{2017}\)
\(=\dfrac{2017}{2017}-\dfrac{1}{2017}\)
\(=\dfrac{2016}{2017}\)
\(B=\dfrac{2013\cdot2015\cdot2017}{2018\cdot2013\cdot\left(2014+1\right)}\)
\(=\dfrac{2013\cdot2015\cdot2017}{2018\cdot2013\cdot2015}\)
\(=\dfrac{2017}{2018}\)
b)
Ta có:
\(A=\dfrac{2016}{2017}=1-\dfrac{1}{2017}\)
\(B=\dfrac{2017}{2018}=1-\dfrac{1}{2018}\)
Vì \(\dfrac{1}{2017}>\dfrac{1}{2018}\)
\(\Rightarrow1-\dfrac{1}{2017}< 1-\dfrac{1}{2018}\)
\(\Rightarrow A< B\)
Vậy \(A< B\).
28 tháng 5 2017
Anh làm nhé!!
Bài làm:
a) Tính A và B
\(A=2\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{2015.2017}\right)\\ =\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2015.2017}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\\ =1-\dfrac{1}{2017}=\dfrac{2016}{2017}\)
\(B=\dfrac{2013.2015.2017}{2018.2013.\left(2014+1\right)}\\ =\dfrac{2013.2015.2017}{2018.2013.2015}=\dfrac{2017}{2018}\)
b) So sánh A và B.
Ta có: \(A=\dfrac{2016}{2017}=1-\dfrac{1}{2017}\\ B=\dfrac{2017}{2018}=1-\dfrac{1}{2018}\\ Mà:\dfrac{1}{2017}>\dfrac{1}{2018}\\ =>1-\dfrac{1}{2017}< 1-\dfrac{1}{2018}\\ =>A< B\)
BP
30 tháng 8 2023
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
14 tháng 4 2017
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 \(\dfrac{A}{B}=\dfrac{1}{2016}\)
11 tháng 12 2021
Áp dụng t/c dtsbn ta có:
\(\dfrac{a}{2b}=\dfrac{2b}{c}=\dfrac{c}{a}=\dfrac{a+2b+c}{2b+c+a}=1\)
\(\dfrac{a}{2b}=1\Rightarrow a=2b\\ \dfrac{2b}{c}=1\Rightarrow c=2b\\ \dfrac{c}{a}=1\Rightarrow a=c\\ \Rightarrow a=2b=c\)
\(M=\dfrac{a^3.c^2.b^{2015}}{b^{2020}}=\dfrac{a^3.a^2}{b^5}=\dfrac{a^5}{b^5}=\dfrac{\left(2b\right)^5}{b^5}=\dfrac{32b^5}{b^5}=32\)
AF
20 tháng 4 2018
A=\(\dfrac{2016.2017+1}{2016.2017}=\dfrac{2016.2017}{2016.2017}+\dfrac{1}{2016.2017}=1+\dfrac{1}{2016.2017}\)
A=\(\dfrac{2017.2018+1}{2017.2018}=\dfrac{2017.2018}{2017.2018}+\dfrac{1}{2017.2018}=1+\dfrac{1}{2017.2018}\)
Mà 1=1; \(\dfrac{1}{2016.2017}\)>\(\dfrac{1}{2017.2018}\) nên A>B
A = 1
B = -1
=> A + B = 0
bai cuoi de hk2 toan lop 6 cua truong minh do