Tính giá trị các biểu thức sau:
a, A = (1 - 1/2) + ( 1 - 1/4 )+ ( 1 - 1/8 ) +...+ (1 - 1/512) + (1 - 1/1024)
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tính biểu thức sau
\(a=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+..........+\frac{1}{512}+\frac{1}{1024}\)
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)
\(2A=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{10}}+\frac{1}{2^{11}}\)
\(2A-A=\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{10}}+\frac{1}{2^{11}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\right)\)
\(A=2^{11}-2\)
(1981 x 1982 - 990) : (1980 x 1982 + 992)
=(1980 x 1982+1982 -990) : (1980 x 1982 +992)
=(1980 x 1982 + 992) : ( 1980 x 1982 + 992)
=1
A = 1/2 - 1/4 - 1/8 -...- 1/512 - 1/1024
2A = 2(1/2 - 1/4 - 1/8 -...- 1/512 - 1/1024)
2A = 1 - 1/2 - 1/8 -...- 1/1024 - 1/2048
2A - A = 1 - 1/2 - 1/8 -....- 1/1024 - 1/2048 - (1/2 - 1/4 - 1/8 - ...- 1/512 - 1/1024)
A = 1 - 1/2048
A = 2047/2048
Em mới học lớp 6, vậy anh thua em rồi. HIHI
Đặt: \(A=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{512}+\dfrac{1}{1024}\)
\(\Rightarrow\dfrac{4}{2}A=\dfrac{4}{2}\left(1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{1024}\right)\)
\(\Rightarrow2A=2+1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{512}\)
\(\Rightarrow2A-A=\left(3+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\right)\)
\(\Rightarrow A=\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(3-1-\dfrac{1}{1024}\right)\)
\(\Rightarrow A=2-\dfrac{1}{1024}\)
\(\Rightarrow A=\dfrac{2047}{1024}\)
1+(1)/(2)+(1)/(4)+(1)/(8)+...+(1)/(512)+(1)/(1024)
A x 2 = 1 - ( 1/2 + 1/4 + 1/8 + 1/16 + ..... + 1/512 + 1/1024 ) - 1/1024
A x 2 = 1 - 1/1024 + A
A x 2 - A = 1 - 1/1024
A = 1 - 1/1024
A = 1023 /1024
\(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(=1-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
đặt A = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
2A = \(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)
2A - A = \(\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
A = \(1-\frac{1}{1024}\)
A = \(\frac{1023}{1024}\)
Thay A vào ta được :
\(1-\frac{1023}{1024}=\frac{1}{1024}\)
Trần Ngọc Mai Anh
Bấm vô đây nhé :
Tính A = 1+1/2+1/4+1/8+…+1/1024
A = ( 1 - \(\dfrac{1}{2}\) ) + ( 1 - \(\dfrac{1}{4}\)) + ( 1 - \(\dfrac{1}{8}\)) +......+ ( 1 - \(\dfrac{1}{512}\)) + ( 1 - \(\dfrac{1}{1024}\))
A = (1 + 1 +....+ 1) - ( \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + ......+ \(\dfrac{1}{512}\) + \(\dfrac{1}{1024}\))
A = ( 1 + 1 +.....+ 1) - ( \(\dfrac{1}{2^1}\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\)+....+ \(\dfrac{1}{2^9}\) + \(\dfrac{1}{2^{10}}\))
Vì trong tổng A có 10 phân số nên
nhóm ( 1 + 1 +....+ 1) có 10 hạng tử là 1
Vậy A = 1 \(\times\) 10 - ( \(\dfrac{1}{2^1}\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\) +..........+ \(\dfrac{1}{2^9}\) + \(\dfrac{1}{2^{10}}\))
Đặt B = \(\dfrac{1}{2^1}\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\) +......+ \(\dfrac{1}{2^9}\) + \(\dfrac{1}{2^{10}}\)
2 \(\times\) B = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\)+........+ \(\dfrac{1}{2^9}\)
2B - B = 1 - \(\dfrac{1}{2^{10}}\)
B = 1 - \(\dfrac{1}{2^{10}}\)
A = 10 + 1 - \(\dfrac{1}{2^{10}}\)
A = 11 - \(\dfrac{1}{2^{10}}\)
em cảm ơn