3/2 + 3/8 + 3/32 + 3/128 + 3/512

= 768/512 + 182/512 + 48/512 + 12/512 + 3/512

= 960/512 + 60/512 + 3/512

= 1023/512 

0,856m=?m

\(...=\dfrac{3}{2}x\left(1+\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{64}+\dfrac{1}{256}\right)\)

\(=\dfrac{3}{2}x\left(\dfrac{256}{256}+\dfrac{64}{256}+\dfrac{16}{256}+\dfrac{4}{256}+\dfrac{1}{256}\right)\)

\(=\dfrac{3}{2}x\dfrac{341}{256}=\dfrac{1023}{512}\)

Đặt tổng trên = A

\(A=\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\)

\(A.4=6+\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}\)

\(A.4-A=\left(6+\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}\right)-\left(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\right)\)

\(A.3=6-\frac{3}{512}=\frac{3069}{512}\)

\(A=\frac{3069}{512}:3=\frac{1023}{512}\)

Đặt A = 3/2 + 3/8 + ... + 3/512
bn tách 
3/2 = 3/2^1
3/8 = 3/2^3
...
3/512 = 3/2^9
Rồi nhân nó lên trừ được bao nhiêu - đi A ban đầu là đc
k nka

=1023/512

C          =        \(\dfrac{3}{2}\) + \(\dfrac{3}{4}\) + \(\dfrac{3}{8}\)+ \(\dfrac{3}{16}\)+...........+\(\dfrac{3}{128}\)

C\(\times\)2      =  3 + \(\dfrac{3}{2}\) + \(\dfrac{3}{4}\) + \(\dfrac{3}{8}\) + \(\dfrac{3}{16}\)+...+\(\dfrac{1}{64}\)

C\(\times\)2 - C =  3 - \(\dfrac{3}{128}\)

C           = \(\dfrac{381}{128}\)

Câu E em xem lại đề nhé 

Đặt tổng trên là: \(A\)

\(A=\dfrac{3}{2}+\dfrac{3}{8}+\dfrac{3}{32}+\dfrac{3}{128}+\dfrac{3}{512}\)

\(\Rightarrow A.4=6+\dfrac{3}{2}+\dfrac{3}{8}+\dfrac{3}{32}+\dfrac{3}{128}\)

\(\Rightarrow A.4-A=\left(6+\dfrac{3}{2}+\dfrac{3}{8}+\dfrac{3}{32}+\dfrac{3}{128}\right)-\left(\dfrac{3}{2}+\dfrac{3}{8}+\dfrac{3}{32}+\dfrac{3}{128}+\dfrac{3}{512}\right)\)

\(\Rightarrow A.3=6-\dfrac{3}{512}=\dfrac{3069}{512}\)

\(\Rightarrow A=\dfrac{3069}{512}:3=\dfrac{1023}{512}\)

3/512 là gì vậy bn?

chào bn

Help

\(P+\frac{1}{512}=\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{4}{512}=\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{4}{128}=\)

\(=\frac{3}{2}+\frac{3}{8}+\frac{4}{32}=\frac{3}{2}+\frac{4}{8}=\frac{4}{2}=2\)

\(\Rightarrow P=2-\frac{1}{512}=\frac{1023}{512}\)

\(P=\frac{3}{2}+\frac{3}{8}+...+\frac{3}{512}\)

\(=3.\left(\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^9}\right)\)

\(4P=3\left(\frac{1}{2^3}+\frac{1}{2^5}+...+\frac{1}{2^{11}}\right)\)

\(4P-P=3\left(\frac{1}{2}-\frac{1}{2^{11}}\right)\)

\(3P=3\left(\frac{1}{2}-\frac{1}{2^{11}}\right)\)

\(P=\frac{1}{2}-\frac{1}{2^{11}}=\frac{2^{10}-1}{2^{11}}\)