tính bằng cách thuận tiện : \(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...+\frac{1}{128}+\frac{1}{256}\)
Tính bằng cách thuận tiện nhất :
\(\frac{1}{2}\)+ \(\frac{1}{4}\) + \(\frac{1}{8}\)+ \(\frac{1}{16}\)+ \(\frac{1}{32}\)+ \(\frac{1}{64}\)+ \(\frac{1}{128}\)
Đặt :
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(\Leftrightarrow\)\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}\)
\(\Leftrightarrow\)\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}\)
\(\Leftrightarrow\)\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^6}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)\)
\(\Leftrightarrow\)\(A=1-\frac{1}{2^7}\)
Vậy \(A=1-\frac{1}{2^7}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{64}-\frac{1}{128}\)
\(=1-\frac{1}{128}\)
\(=\frac{127}{128}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{64}-\frac{1}{128}\)
= \(1-\frac{1}{128}\)
= \(\frac{127}{128}\)
Tính : \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
bài 7:tính bằng cách thuận tiện nhất
a, 1/3+2/3...+3\(\frac{1}{3}\) + 3\(\frac{2}{3}\) =.......
b,9+9/2+9/4+9/8......+9/128+9/256=......
c,\(\frac{5}{3}\) + \(\frac{5}{3\cdot5}\) + \(\frac{5}{5\cdot7}\) + ......... +\(\frac{5}{101\cdot103}\) =.........
Tính :
B = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+......+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
Theo đề bài ta có :
\(2B=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}\)
\(\Leftrightarrow2B-B=\left(1+\frac{1}{2}+...+\frac{1}{128}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\right)\)
\(\Leftrightarrow B=1-\frac{1}{256}\)
\(\Leftrightarrow B=\frac{255}{256}\)
\(B=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+..+\frac{1}{256}\)
\(\Rightarrow B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..+\frac{1}{2^8}\)
\(\Rightarrow2B=1+\frac{1}{2}+\frac{1}{2^2}+..+\frac{1}{2^7}\)
\(\Rightarrow2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\right)\)
\(\Rightarrow B=1-\frac{1}{2^8}\)
p/s: bài lớp 6!
\(B=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(B=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}+\frac{1}{2^8}\)
\(\Rightarrow2B=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}\)
\(\Rightarrow2B-B=1-\frac{1}{2^8}\)
\(B=1-\frac{1}{2^8}\)
tính bằng cách thuận tiện nhất: \(6\frac{2}{7}+7\frac{3}{5}+8\frac{6}{9}+9\frac{1}{4}+\frac{2}{5}+\frac{5}{7}+\frac{1}{3}+\frac{3}{4}\)
Tính nhanh:
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(A\cdot2=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...+\frac{1}{256}\right)\cdot2\)
\(=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...+\frac{1}{128}\)
\(A\cdot2-A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...+\frac{1}{128}-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\right)\)
\(A=1-\frac{1}{256}=\frac{255}{256}\)
\(A=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^7}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^7}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)\)
\(A=1-\frac{1}{2^8}\)
\(A=\frac{2^8-1}{2^8}\)
\(A=\frac{255}{256}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{128}-\frac{1}{256}\)
\(A=1-\frac{1}{256}\)
\(A=\frac{255}{256}\)
b =\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}...+\frac{1}{128}+\frac{1}{256}\)
giong bai kia thui
nhung bay gio mk phai off rui
6.30 mk se giai cho nha
chuc bn hoc tot
$$$
Số thích hợp để điền vào chỗ chấm là:397/384
k mk nhé!thanks
\(\frac{1}{2}=1-\frac{1}{2}\)
\(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\)
\(\frac{1}{8}=\frac{1}{4}-\frac{1}{8}\)
\(\frac{1}{16}=\frac{1}{8}-\frac{1}{16}\)\(......\)
Vay B=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+......+\frac{1}{64}-\frac{1}{128}+\frac{1}{128}-\frac{1}{256}\)
\(B=1-\frac{1}{256}\)
\(B=\frac{255}{256}\)
Tính:
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
Yêu cầu lời giải đầy đủ
Đặt A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512
2A = 1/2 x 2 + 1/4 x 2 + 1/8 x 2 + 1/16 x 2 + 1/32 x 2 + 1/64 x 2 + 1/128 x 2 + 1/256 x 2 + 1/512 x 2
2A = 1 + 1/2 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256
2A - A = ( 1 + 1/2 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 ) - ( 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512 )
A = 1 - 1/512
A = 511/512
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{128}-\frac{1}{256}\)
=\(1-\frac{1}{256}\)
=\(\frac{255}{256}\)
1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256
= 128/256 + 64/256 + 32/256 + 16/256 + 8/256 + 4/256 + 2/128 + 1/256
= 255/256