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NP
3
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
1 tháng 12 2023
A = \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\)
2 \(\times\) A = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\)
2 \(\times\) A - A = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) - (\(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\))
A = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) - \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) - \(\dfrac{1}{8}\) - \(\dfrac{1}{16}\) - \(\dfrac{1}{32}\)
A = 1 - \(\dfrac{1}{32}\)
A = \(\dfrac{31}{32}\)
HT
22 tháng 10 2020
bài 1 tính nhanh
mik xin sửa đề câu a thành thế này ~
\(a,\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(A\cdot2=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(A\cdot2-A=\) ( \(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\) ) - ( \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\) )
\(A=1-\frac{1}{256}\)
\(A=\frac{255}{256}\)
\(b,\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
đặt \(B=\) \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(B\cdot3=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(B\cdot3-B=\) ( \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)) - \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\) )
\(B\cdot2=\) \(1-\frac{1}{729}\)
\(B\cdot2=\frac{728}{729}\)
\(B=\frac{728}{729}:2\)
\(B=\frac{364}{729}\)
\(c,\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)
ĐẶT \(C=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)
\(C=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\)
\(C=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(C=\frac{1}{1}-\frac{1}{6}\)
\(C=\frac{5}{6}\)
SB
3
16 tháng 5 2015
1+ 1 /3+1/9+1/27+1/81+1/243+1/729.
Đặt:
S = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243
Nhân S với 3 ta có:
S x 3 = 3 +1+ 1/3 + 1/9 + 1/27 + 1/81
Vậy:
S x 3 - S = 3 - 1/243
2S = 728/243
S = 364/243
tick đúng nha
NN
1
DD
Đoàn Đức Hà
Giáo viên
16 tháng 10 2021
\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(3\times A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(3\times A-A=\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)-\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\right)\)
\(2\times A=1-\frac{1}{729}=\frac{728}{729}\)
\(A=\frac{364}{729}\)
16 tháng 9 2018
\(A=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2048}\)
\(A=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+...+\left(\frac{1}{1024}-\frac{1}{2048}\right)\)
\(A=1-\frac{1}{2048}\)
\(\Rightarrow\)\(A=\frac{2047}{2048}\)
\(3B=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(3B-B=1-\frac{1}{2187}\)
\(2B=\frac{2186}{2187}\)
\(\Rightarrow B=\frac{2186}{4374}=\frac{1093}{2187}\)
BN
20 tháng 10 2017
a)
\(\frac{32+16+8+4+2+1+128}{64}\)
\(\frac{191}{64}\)
B)
\(\frac{81+27+9+3+1+243}{243}\)
\(\frac{364}{243}\)
Mình lười làm qua :(
tính nhanh p/s 1+ 5/4 + 5/8 + 5/16 + 5/32 + 5/64
9 tháng 5 2023
b: A=1/3+1/9+...+1/3^10
=>3A=1+1/3+...+1/3^9
=>A*2=1-1/3^10=(3^10-1)/3^10
=>A=(3^10-1)/(2*3^10)
c: C=3/2+3/8+3/32+3/128+3/512
=>4C=6+3/2+...+3/128
=>3C=6-3/512
=>C=1023/512
d: A=1/2+...+1/256
=>2A=1+1/2+...+1/128
=>A=1-1/256=255/256
8 tháng 8 2017
a) = \(\frac{127}{96}\)
b) = \(\frac{255}{256}\)
c) Mik bỏ nha
d) = \(\frac{1023}{512}\)
e) = \(\frac{2343}{625}\)
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}\)
\(A=1-\frac{1}{64}\)
\(A=\frac{63}{64}\)
\(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(3B=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)
\(3B-B=1-\frac{1}{243}\)
\(2B=\frac{242}{243}\)
\(B=\frac{242}{243}\div2\)
\(B=\frac{121}{243}\)
a.A=1/2+1/4+1/8+1/16+1/32+1/64
A= \(\frac{1}{1\cdot2}+\frac{1}{2\cdot2}+\frac{1}{2\cdot4}+\frac{1}{4\cdot4}+\frac{1}{4\cdot8}+\frac{1}{8\cdot8}\)
= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{8}\)
= 1 - 1/8 = 7/8
b.B=1/3+1/9+1/27+1/81+1/243
B= \(\frac{1}{1\cdot3}+\frac{1}{3\cdot3}+\frac{1}{3\cdot9}+\frac{1}{9\cdot9}+\frac{1}{9\cdot27}\)
= 1 - 1/27 = 26/27