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A = \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{64}\) + \(\dfrac{1}{128}\) + \(\dfrac{1}{256}\)
2A = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{64}\) + \(\dfrac{1}{128}\)
2A - A = 1 - \(\dfrac{1}{256}\)
A = \(\dfrac{255}{256}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
X x (1/2+1/4+1/8+1/16+1/32+1/64+1/128) = 127/128
X x 127/128 = 127/128
X = 127/128 : 127/128
X = 1
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\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(2A=2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\right)\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\right)\)
\(-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\right)\)
\(\Rightarrow A=1-\frac{1}{256}\)
\(\Rightarrow A=\frac{255}{256}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: 4A=4+4^2+...+4^9
=>3A=4^9-1
=>A=(4^9-1)/3
b: 2A=1+1/2+...+1/2^7
=>A=1-1/256=255/256
c: =1-1/5+1/5-1/9+...+1/85-1/89
=1-1/89=88/89
d: =1/3(3/1*4+3/4*7+...+3/304*307)
=1/3(1-1/4+1/4-1/7+...+1/304-1/307)
=1/3*306/307=102/307
e: E=1-1/2+1/2-1/3+...+1/11-1/12
=1-1/12=11/12
g: =2/5(1-1/6+1/6-1/11+...+1/96-1/101)
=2/5*100/101=40/101
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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
![](https://rs.olm.vn/images/avt/0.png?1311)
đề phải là 1 +1/2 + 1/4 +1/32 + 1/64 + 1/128 +1/256 +/512 +1/1024 moi dug
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+......+\frac{1}{128}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+........+\frac{1}{64}\)
Lấy 2A-A ta có:
2A-A=\(\left(1+\frac{1}{2}+\frac{1}{4}+....+\frac{1}{64}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\right)\)
\(\Rightarrow A=1-\frac{1}{128}\)
\(\Rightarrow A=\frac{127}{128}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/2+1/4+1/8+1/16+....+1/128
=1-1/2+1/2-1/4+1/4-1/8+1/8-1/16+...+1/64-1/128
=1-1/128
=127/128
![](https://rs.olm.vn/images/avt/0.png?1311)
vì quá dễ nên mình không thể trả lời bạn được nhé!
![](https://rs.olm.vn/images/avt/0.png?1311)
Ơ bài này toán 6 nhé:
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+..+\frac{1}{128}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\)
\(2A=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)\)
\(A=1-\frac{1}{2^7}\)
Vậy \(A=1-\frac{1}{2^7}\)
1/2 + 1/4 +1/8 +....1/128
=1 - 1/2 + 1/2 - 1/4 + 1/4 -....+ 1/64 - 1/128
=1 - 1/128
=127/128
Học tốt!