So sánh A vớiB
a, A=1+2+3+...+1000
B=1×2×....×11
b, A=1×2×....×20
B=1+2+....+10^6
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NC
28 tháng 8 2020
Bài làm:
Ta có: \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}\)
\(A=\left(1+\frac{1}{3}+...+\frac{1}{9}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\)
\(A=\left[\left(1+\frac{1}{3}+...+\frac{1}{9}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\right]-\left[\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\right]\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)=B\)
Vậy A = B
16 tháng 6 2017
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(A=\left(1+\frac{1}{3}+...+\frac{1}{9}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}+\frac{1}{10}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}+\frac{1}{10}\right)-\left(1+\frac{1}{2}+...+\frac{1}{5}\right)\)
\(A=\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+...+\frac{1}{10}\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{5}\right)\)
Vậy A = B và A = 1/6 + 1/7 + 1/8 + 1/9 + 1/10
TH
16 tháng 6 2017
1/ A= \(\left(\frac{1}{1.2}\right)+\left(\frac{1}{3.4}\right)+...+\left(\frac{1}{9.10}\right)\)
B=(1/1+1/2+1/3+...+1/10)- (1/1+1/2+...+1/5)
<=> B=1/6+1/7+1/8+1/9+1/10.
DH
0
CB
0
HT
1
18 tháng 4 2016
=1/2.(1+1/3+1/6+...+1/45)
=1/2+1/6+1/12+1/20+...+1/90
=1/1.2+1/2.3+1/3.4+...+1/9.10
=(1/1-1/2)+(1/2-1/3)+...+(1/9-1/10)
=1-1/10
=9/10
dung do..k hegg