A=1-1/2+1/3-1/4+1/5-...-1/2018+1/2019
và B=1/1010+1/1011+...+1/2019
Tính ( A-B-1)^2019
KT
Những câu hỏi liên quan
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
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+...-1/2018+1/2019
B=1/1010+1/1011+...+1/2018+1/2019
Tính (A-B)2020
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
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 A = 1 - \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\)
và B = \(\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}+\frac{1}{2019}\)
Tính \(\text{(A - B - 1) }^{2019}\)
Ta có : \(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2019}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2018}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2018}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1009}\right)\)
\(=\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2019}=B\)
\(\Rightarrow A-B-1=-1\)
\(\Rightarrow\left(A-B-1\right)^{2019}=-1\)
cho A= 1011(1 +1/3+1/5+....+1/2019) và B = 1010(1/2+1/4+1/6+...+1/2020)
so sanh A và B
Cho A = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\)
B = \(\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}+\frac{1}{2019}\)
Tính ( A - B - 1)2019
Ơ !!! Bài này giống bài 5 môn Toán thi cuối học kỳ 2 trường mình nè !!!
Đúng 0
Bình luận (0)
Bài làm
Ta có: \(A=\) \(1\) \(-\)\(\frac{1}{2}\)\(+\)\(\frac{1}{3}\)\(-\)\(\frac{1}{4}\)\(+\)\(......\)\(+\)\(\frac{1}{2017}\)\(-\)\(\frac{1}{2018}\)\(+\)\(\frac{1}{2019}\)
\(\Rightarrow\) \(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}-2\left(\frac{1}{2}+\frac{1}{4}+......+\frac{1}{2018}\right)\)
\(\Rightarrow A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{2019}-\left(1+\frac{1}{2}+......+\frac{1}{1009}\right)\)
\(\Rightarrow A=\frac{1}{1010}+\frac{1}{1011}+......+\frac{1}{2019}\)
\(\Rightarrow A=B\)
Khi đó: (A - B - 1)2019 = -12019 = -1
Chúc bạn học tốt. K cho mk nhé! Thank you.
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
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
Đúng 0
Bình luận (0)