A=(1/1009+1/1010+...+1/2016+1/2017).(1-1/2+1/3-1/4+...+1/2015-1/2016)
Những câu hỏi liên quan
A=(1/1009+1/1010+...+1/2016+1/2017)(1-1/2+1/3-1/4+...+1/2015-1/2016)
Tính A=(1/1009+1/1010+...+1/2016+1/2017)(1-1/2+1/3+1/4+...+1/2015+1/2016)
cho S=1-1/2+1/3-1/4+......+1/2015-1/2016+1/2017 và P=1/1009+1/1010+.....+1/2016+1/2017
Tính (S-P)^2016
cho M=1-1/2+1/3-1/4+...+1/2015-1/2016+1/2017
N= 1/1009+1/1010+....+1/2016+1/2017
tính (M-N)^2017
\(M=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}+\dfrac{1}{2017}\)
\(M=\left(1+\dfrac{1}{3}+...+\dfrac{1}{2015}+\dfrac{1}{2017}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2016}\right)\)\(M=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2016}\right)\)\(M=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2008}\right)\)
\(M=\dfrac{1}{2009}+\dfrac{1}{2010}+...+\dfrac{1}{2016}+\dfrac{1}{2017}=N\)
Vậy \(\left(M-N\right)^{2017}=0\)
Đúng 0
Bình luận (0)
A=(1/1009+1/1010+...+1/2015+1/2016):(1/1-1/2+1/3-1/4+...+1/2015-1/2016)
Xét số chia: 1-\(\frac{1}{2}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\) +...+\(\frac{1}{2015}\) - \(\frac{1}{2016}\)
= (1+\(\frac{1}{2}\) + \(\frac{1}{3}\) + \(\frac{1}{4}\) +...+\(\frac{1}{2015}\) + \(\frac{1}{2016}\)) - 2.(\(\frac{1}{2}\) + \(\frac{1}{4}\) + ... + \(\frac{1}{2016}\))
= (1+\(\frac{1}{2}\) + \(\frac{1}{3}\) + \(\frac{1}{4}\) +...+\(\frac{1}{2015}\) + \(\frac{1}{2016}\)) - (1+\(\frac{1}{2}\) + \(\frac{1}{3}\) + \(\frac{1}{4}\) +...+\(\frac{1}{1007}\) + \(\frac{1}{1008}\))
=\(\frac{1}{1009}\) + \(\frac{1}{1010}\) + ... + \(\frac{1}{2015}\)+ \(\frac{1}{2016}\) => A=1
Đúng 0
Bình luận (0)
Tính : A=(1/1009+1/1010+...+1/2015+1/2016):(1/1-1/2+1/3-1/4+...+1/2015+1/2016)
\(A=\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}+\frac{1}{2017}\right): \)\(\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\right)\)
\(\text{đặt}k=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\)
\(K=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(K=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{1008}\right)\)
\(K=\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+....+\frac{1}{2017}\Rightarrow A=1\)
Đúng 0
Bình luận (0)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\)
\(B=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}+\frac{1}{2017}\)
Tính \(\left(A-B\right)^{2016^{2017}}\)
A=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\)
B=\(\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2016}+\frac{1}{2017}\)
tính (A-B)^2017