cho a=1/2+1/3+1/4+...+1/2018 và b=1/2017+2/2016+3/2015+...+2017/1. Tính a/b
Những câu hỏi liên quan
Tính :A= [(2018/1)+(2017/2)+(2016/3)+(2015/4)+...+(4/2015)+(3/2016)+(2/2017)+(1/2018)]/[(2019/1)+(2019/2)+(2019/3)+(2019/4)+...+(2019/2015)+(2019/2016)+(2019/2017)+(2019/2018)+(2019/2019)]
cho a=1/2+1/3+1/4+...+1/2018 và b=1/2017+2/2016+3/2015+...+2017/1. Tính a/b.
Giúp mk với mk cần gấp lắm
Tính :A= [(2018/1)+(2017/2)+(2016/3)+(2015/4)+...+(4/2015)+(3/2016)+(2/2017)+(1/2018)]/[(2019/2)+(2019/3)+(2019/4)+(2019/5)+...+(2019/2015)+(2019/2016)+(2019/2017)+(2019/2018)+(2019/2019)]
A= ( 1/2017+ 2/2016+ 3/2015+...+ 2015/3+ 2016/2+ 2017) : ( 1/2+1/3+1/4+...+1/2017+1/2018)
So sánh:
a) A = 102016 - 2 / 102017 - 2 và B = 202015 + 1 / 102016 + 1
b) A = 20162017 - 3 / 20162018 - 3 và B = 20162016 + 3 / 20162017 + 3
c) A = 20172016 - 2015 / 20172017 - 2015 và B = 20172015 + 1 / 20172016 + 1
Cho A=1/2018+2/2017+3/2016+...+2017/2+2018
B=1/2+1/3+1/4+....+1/2019
Tính A/B
\(A=\frac{1}{2018}+\frac{2}{2017}+...+\frac{2017}{2}+2018\)
\(=\left(\frac{1}{2018}+1\right)+\left(1+\frac{2}{2017}\right)+...+\left(\frac{2017}{2}+1\right)+1\)(2018 số hạng 1)
\(=\frac{2019}{2018}+\frac{2019}{2017}+...+\frac{2019}{2}+\frac{2019}{2019}=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)\)
Mà \(B=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\)
=> Khi đó : \(\frac{A}{B}=\frac{2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}}=2019\)
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Tính tỉ số A/B biết:
A=1/2 + 1/3 + 1/4 + ... + 1/2017 + 1/2018 + 1/2019
B=2018/1 + 2017/2 + 2016/3 + ... + 2/2017 + 1/2018
\( S =1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\)
\(\Rightarrow S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1} {2019}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right) \)
\(\Rightarrow S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)
\(\(\Rightarrow S=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2019}\) \(\Rightarrow S=P\)\)
\(B=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{1}{2018}\)
\(B=1+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{1}{2018}+1\right)\)
\(B=\frac{2019}{2019}+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2018}\)
\(B=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)\)
ta có \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}}{2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)}=\frac{1}{2019}\)
Cho
A = 1/2 + 1/3 + 1/4 + ... + 1/2017
B = 1/2016 + 2/2015 +3/2014+ ...+ 2015/2 + 2016/1
Tính B : A
Ta có: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)
Đúng 1
Bình luận (0)
Tính:
A=2019/2018 - 2018/2017 + 2017/2016 - 2016/2015
B=1/2019 - 1/2018 + 1/2017 - 1/2016
C=1/2017 - 1/2016 + 1/2015 - 1/2014