\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{15}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{14}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{14}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{15}}\right)\)
\(A=1-\frac{1}{2^{15}}\)\
\(A=1-\frac{1}{32768}\)
\(A=\frac{32767}{32768}\)
A = 2mu 0 + 2 mu 1 + 2mu 2 + ...+ 2 mu 50
A = 2mu 0 + 2 mu 1 + 2mu 2 + ...+ 2 mu 50
cho a = 1+2+ 2 mũ 2 + 2 mũ 3 + 2mu 4 +....+ 2 mu 200 hay việt á công 1 dưới dạng lũy thừa
tim so tu nhien n,biet: a) (2mu n+1)=27 b) (n-2)mu 2=(n-2)mu 4 c)3 mu n+1=81 d)2mu2n+1=512
bai1 12mu n+1 + 11mu n+2 chia hết cho 133
bai2 3mu n+2 -2mu n+2 + 3 mu n -2mu n
(5 mu 4 + 4 mu 7) (8 mu9 - 2mu 7) (2mu 4-4 mu 2)
bai1
5mu3+3mu5
bai2
(x-1)mu3=125
720/(41-(2*x-5))=2mu 3*5
bai3
1 phan 9 * 3 mu 4 * 3 mu n =3 mu 7
(2 mu 2 chia 4)* 2 mu n = 4
bai4
2 * 2mu2 * 2mu3 * 2mu4 *.......*2 mu100
cho a= 2mu 2+ 2 mu 3+2mu4+....+2mu 12
chung to rang a chia hết cho 39
1) hai mu2+2 mu3 +2mu10 .......+2mu100
2) tim chu so tan cungcua so[2mu 2013]o tren co mu 2
