A=\(\frac{2.2012}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+..+2018}}\)
B=2018
so sánh A và B
K
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17 tháng 12 2017
Ta có:
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}-1-\frac{1}{2}-...-\frac{1}{1009}\)
\(=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}=B\)
14 tháng 5 2019
B= 1/1.2+1/2.3+...+1/2019.2020
B=1/1-1/2+1/2-1/3+...+1/2019-1/2020
B=1-1/2020=2020/2020-1/2020=2019/2020
22 tháng 5 2019
đặt 22018 = a ; 32019 = b ; 52020 = c
Ta có : \(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\)
\(B=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(2B=\frac{2}{1.2}+\frac{2}{3.4}+...+\frac{2}{2019.2020}\)
\(< 1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}+\frac{1}{2019.2020}\)
\(2B< 1+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{2019-2018}{2018.2019}+\frac{2020-2019}{2019.2020}\)
\(2B< 1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}=1+\frac{1}{2}-\frac{1}{2020}< 1+\frac{1}{2}\)
\(B< \frac{3}{4}\)
\(\Rightarrow A>1>\frac{3}{4}>B\)
22 tháng 5 2019
Mình chỉ biết cách tính B thôi, đây nhé:
B= \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2019.2020}\)
B=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(B=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2019}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2020}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2020}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-2\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1010}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1010}\right)\)
\(B=\frac{1}{1011}+\frac{1}{1012}+....+\frac{1}{2019}+\frac{1}{2020}\)
14 tháng 5 2019
\(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1.\)
Với : \(a=2^{2018};.b=3^{2019};,c=5^{2020}.\)
Và : \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2019.2020}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\Leftrightarrow\)
\(B=1-\frac{1}{2020}< 1< A\)
23 tháng 8 2018
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)..........\left(\frac{1}{2018^2}-1\right)\)
Ta có :
\(\frac{1}{2^2}-1>-\frac{1}{2}\)
\(\frac{1}{3^2}-1>-\frac{1}{2}\)
...........
\(\frac{1}{2018^2}-1>\frac{1}{2}\)
\(\Rightarrow A>B\)
14 tháng 5 2018
Ta có : \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}\)
\(B=75\%=\frac{3}{4}\)
Ta có : \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}\)
\(=\frac{1}{4}+\left(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}\right)< \frac{1}{4}+\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\right)\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{2018}=\frac{3}{4}-\frac{1}{2018}< \frac{3}{4}\)
\(\Rightarrow A< B\)
14 tháng 5 2018
Ta có : \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}\)
\(B=75\%=\frac{3}{4}\)
Ta có : \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}\)
\(=\frac{1}{4}+\left(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}\right)< \frac{1}{4}+\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\right)\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{2018}=\frac{3}{4}-\frac{1}{2018}< \frac{3}{4}\)
\(\Rightarrow A< B\)
14 tháng 5 2018
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2017^2}-1\right)\left(\frac{1}{2018^2}-1\right)\)
\(A=\frac{\left(1-2^2\right)\left(1-3^2\right)\left(1-4^2\right)...\left(1-2018^2\right)}{2^23^24^2...2018^2}\)
\(A=\frac{-1\cdot3\cdot\left(-2\right)\cdot4\cdot\left(-3\right)\cdot5\cdot...\cdot\left(-2016\right)\cdot2018}{2018!^2}\)
\(A=\frac{2016!\cdot3\cdot4\cdot5\cdot...\cdot2018}{2018!^2}=\frac{2016!\cdot2018!}{2018!^2\cdot2!}=\frac{2016!}{2!2018!}=\frac{1}{2!\cdot2017\cdot2018}>0>-\frac{1}{2}=B\)
7 tháng 9 2018
A = (1/2+1)(1/2-1)(1/3+1)(1/3-1)....(1/2018+1)(1/2018-1) đặt các tích phần tử có dấu + là X, tích các phần tử có dấu - là Y => A= X.Y
X = 3/2.4/3.5/4.....2019/2018 = 2019/2
Y= (-1/2)(-2/3)(-3/4)...(-2017/2018) = -1/2018 (tích của 2017 số <0)
A= X.Y = -2019/2018.1/2 < B= -1/2