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gọi biểu thức là A
A=1/2+1/4+1/8+...+1/2048=1/2+1/2^2+1/2^3+...+1/2^10
=>2A=1+1/2+1/2^2+...+1/2^9
=>A=2A-A(bạn đặt cột dọc ra rồi sẽ thấy:1/2-1/2=0;1/2^2-1/2^2=0;...)Ta được kết quả bằng 1+1/2^10
Đặt A =1/2 + 1/4 + 1/8 + ...+ 1/1024 + 1/2048
A= 1/2 + 1/2^2 + 1/2^3+...+ 1/2^10 + 1/2^11
2A= 1 +1/2 + 1/2^2 +...+ 1/2^9 + 1/2^10
2A-A= (1 +1/2 + 1/2^2 +...+ 1/2^9 + 1/2^10) - (1/2 + 1/2^2 + 1/2^3+...+ 1/2^10 + 1/2^11)
A= 1+1/2 + 1/2^2 +...+ 1/2^9 + 1/2^10 - 1/2 - 1/2^2 - 1/2^3 - ...- 1/2^10 - 1/2^11
A= 1- 1/2^11
A= 2047/ 2048
Đặt A = \(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}\)
\(\Rightarrow2A=2-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}\)
\(\Rightarrow2A-A=1-1+\frac{1}{16}\)
\(\Rightarrow A=\frac{1}{16}\)
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
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{2048}\)
\(=1-\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2048}\)
\(=1-\frac{1}{2048}\)
\(=\frac{2047}{2048}\)
Ta có :\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}+\frac{1}{2048}\)
nên \(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}+\frac{1}{512}+\frac{1}{1024}\)
Do đó : \(2.A-A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}+\frac{1}{512}+\frac{1}{1024}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}+\frac{1}{2048}\right)\)
\(A=1-\frac{1}{2048}=\frac{2047}{2048}\)
Nhớ k giùm mình nhớ
\(\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}\right):\left(\frac{1}{6}+\frac{1}{10}-\frac{1}{15}\right)\)
=\(\frac{1}{3}+\frac{1}{5}\)
=\(\frac{5}{3}\)
= 2047/2048 nha bạn k mình nha
1024 phan 2048
A=1/2+1/4+1/8+1/16+...+1/2048
2A=1+1/2+1/4+1/1/8+...+1/1024
2A-A=(1+1/2+...+1/1024)-(1/2+1/4+...+1/2048)
A=1-1/2048
A=2047/2048
Gọi biểu thức trên là A
A=1/2+1/2^2+1/2^3+....+1/2^10+1/2^11
2A=1+1/2+1/2^2+...+1/2^11+1/2^12
2A-A=1/2^12+1
A=1/2^12+1
(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
trả lời
=1
chúc bn
hc tốt
trả lời
=1
chúc bn
hc tốt
trả lời
=1
~ Học tốt ~
Trả lời:
=1
#Nhi.
***
Trả lời:
= ko
~ Học tốt nhá ~
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}+\frac{1}{2048}\)
\(A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}+\frac{1}{2^{11}}\)
\(\Rightarrow2A=\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)
\(\Rightarrow2A-A=\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^9}+\frac{1}{2^{10}}-\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{10}}-\frac{1}{2^{11}}\)
\(\Rightarrow A=\frac{1}{2^0}+\frac{1}{2^1}-\frac{1}{2^1}+\frac{1}{2^2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^3}...+\frac{1}{2^{10}}-\frac{1}{2^{10}}-\frac{1}{2^{11}}\)
\(\Rightarrow A=\frac{1}{1}+0+0+...+0-\frac{1}{2^{11}}\)
\(\Rightarrow A=1-\frac{1}{2^{11}}\)\(A=1-\frac{1}{2^{11}}\)
Vậy
đáp án là:1
trả lời :
=1
trả lời bằng 1 chúc bn hc giỏi
Sỹ
hoi dai thê
đáp số là 1 nha bạn
bang 1
hoctot nhe ban ok
Đáp án là 1