Cho A = 1 - 1/2 + 1/3 + 1/4 + .... + 1/2017 +1/2018 + 1/2019
B = 1/1010 + 1/1011 + 1/1012 + .... + 1/2017 + 1/2018 + 1/2019
Tính (A-B-1)2019
Cho A = 1 - 1/2 + 1/3 + 1/4 + .... + 1/2017 +1/2018 + 1/2019
B = 1/1010 + 1/1011 + 1/1012 + .... + 1/2017 + 1/2018 + 1/2019
Tính (A-B-1)2019
Cho A=1-1/2+1/3-1/4+........+1/2017-1/2018 và
B=1/1010+1/1011+1/1012+.......+1/2017+1/2018
Tính A/B^2018
ta có: \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(A=\left(1+\frac{1}{3}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)
\(A=\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2017}+\frac{1}{2018}\)
\(\Rightarrow A=B\left(=\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2017}+\frac{1}{2018}\right)\)
\(\Rightarrow\frac{A}{B^{2018}}=\frac{A}{A.B^{2017}}=\frac{1}{B^{2017}}\)
=> \(\frac{A}{B^{2018}}=\frac{1}{\left(\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2017}+\frac{1}{2018}\right)^{2017}}\)
cho A= 1-1/2+1/3-1/4+.....-1/2018+1/2019 và B=1/1010+1/1011+...+1/2018+1/2019
Tính (A+B) mũ 2020
A=1-\(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}+\dfrac{1}{2019}\)
B=\(\dfrac{1}{1010}+\dfrac{1}{1011}+\dfrac{1}{1012}+...+\dfrac{1}{2019}\)
Tính \(^{\left(A-B\right)^{2019}}\)
cho S =1-1/2+1/3-1/4+...+1/2017-1/2018+1/2019
Va P=1/1010+1/1011+1/1012+...+1/2019
Tinh (S-P)2018
LM GIÚP MÌNH VỚI
CM ƠN TRƯỚC NHA
Ta có : S =\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\)
\(\Rightarrow S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\)\(-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(\Rightarrow S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\)\(-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)
\(\Rightarrow S=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2019}\)
\(\Rightarrow S=P\)
Khi đó : \(\left(S-P\right)^{2018}=0^{2018}=0\)
k chi mik nha!
-.-
A=1+1/2+1/3+1/4+...+1/2^2018-1 Chứng tỏ A<2018
Cho A =\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{\text{4}}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\)
và B=\(\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2018}+\frac{1}{2019}\)
Tính \(\left(A-B-1\right)^{2019}\)
A=(1+1/3+...+1/2019)-(1/2+1/4+...+1/2018)
A=(1+1/3+...+1/2019)+(1/2+1/4+...+1/2018)-(1/2+1/4+...+1/2018).2
A=(1+1/2+1/3+1/4+...+1/2019)-(1+1/2+...+1/1009)
A=1/1010+1/1011+...+1/2019
=) A=B
=) (A-B-1)^2019=-1
tính: D=(1+1/3+1/5+...+1/2017)-(1/2+1/4+1/6+...+1/2018)/1/1010+1/1011+1/1012+...+1/2018
Bạn có thể viết lại đề theo phân số như thế này được không \(\frac{7}{12}\)bạn viết thế mk ko hiểu
Bn viết lại đề nhanh mk làm cho
Chúc bn học tốt
Tính:
E=(1+1/3+1/5+...+1/2017)-(1/2+1/4+1/6+...+1/2018) / 1/1010+1/1011+1/1012+...+1/2018
So sánh A và B biết :
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(B=\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2018}\)
Ta có:
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}-1-\frac{1}{2}-...-\frac{1}{1009}\)
\(=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}=B\)
Nhanh lên giúp mình với !
Ngày mai mình phải nộp rồi.
Làm nhanh nhất, đúng nhất, rõ ràng mình k cho