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\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}=\frac{32}{64}+\frac{16}{64}+\frac{8}{64}+\frac{4}{64}+\frac{2}{32}+\frac{1}{64}\)
\(\frac{32+16+8+4+2}{64}=\frac{62}{64}=\frac{31}{32}\)
Tk mh nhé , mơn nhìu !!!
~ HOK TỐT ~
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)\(+\frac{1}{64}\)
= 32/64 + 16/64 + 8/64 + 4/64 + 2/64 + 1/64
= 63/64
Ta có: \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}\)
=>2A=\(1+\frac{1}{2^2}+...+\frac{1}{2^4}+\frac{1}{2^5}\)
=>2A-A=(\(1+\frac{1}{2^2}+...+\frac{1}{2^4}+\frac{1}{2^5}\))--(\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}\))
=>A=\(1-\frac{1}{2^6}\)
=>A=\(\frac{63}{64}\)
a) 2003x14+1988+2001+2002 / 2002+2002x503+504x2002
=(2002+1)x14+1988+2001+2002 / 2002x(503+1+504)
=2002x14+(14+1988)+2002+2001 / 2002x1008
=2002x(14+1+1)+2001 / 2002x1008
đến đoạn nay mk thấy đề có ì đó sai sai rồi đó, 2001 đáng lẽ phải bằng 2002 mới đúng chứ, đề ko lỗi thì cho mk xin lỗi nha
b) 1/4 + 1/8 + 1/16 + 1/64 + 1/128 (bạn thiếu 100 ở đầu mẫu)
gọi tổng sau là a, ta có
A = 1/2^2 + 1/2^3 + 1/2^4 + 1/2^5 + 1/2^6 + 1/2^7
2xA = 1/2 + 1/2^2 + 1/2^3 + 1/2^4 + 1/2^5 + 1/2^6
2xA-A = (1/2 + 1/2^2 + 1/2^3 + 1/2^4 + 1/2^5 + 1/2^6) - (1/2^2 + 1/2^3 + 1/2^4 + 1/2^5 + 1/2^6 + 1/2^7)
A = 1/2 - 1/2^7
A = 2^6-1/2^7
chúc bạn học tốt nha
Ta có: \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
= \(\frac{1}{2}+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+....+\left(\frac{1}{64}-\frac{1}{128}\right)\)
=\(\frac{1}{2}+\frac{1}{2}-\frac{1}{128}\)
\(=1-\frac{1}{128}=\frac{127}{128}\)
Cách 1:
B=1/2+1/4+1/8+1/16+1/32+1/64
B=1-1/2 + 1/2-1/4 + 1/4-1/8 +1/8-1/16 + 1/16-1/32 + 1/32-1/64
B=1-1/64
B=63/64
Cách 2:
B=1/2+1/4+1/8+1/16+1/32+1/64
B=1/21+1/22+1/23+1/24+1/25+1/26
2B=1+1/21+1/2^2+1/2^3+1/2^4+1/2^5
2B-B=1-1/2^6
B=1-1/64
B=63/64
Đặt A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64
2A = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32
2A - A = (1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32) - (1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64)
A = 1 - 1/64
A = 63/64
Theo đề bài ta có :
\(2B=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}\)
\(\Leftrightarrow2B-B=\left(1+\frac{1}{2}+...+\frac{1}{128}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\right)\)
\(\Leftrightarrow B=1-\frac{1}{256}\)
\(\Leftrightarrow B=\frac{255}{256}\)
\(B=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+..+\frac{1}{256}\)
\(\Rightarrow B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..+\frac{1}{2^8}\)
\(\Rightarrow2B=1+\frac{1}{2}+\frac{1}{2^2}+..+\frac{1}{2^7}\)
\(\Rightarrow2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\right)\)
\(\Rightarrow B=1-\frac{1}{2^8}\)
\(ĐặtA=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)
\(A=1-\frac{1}{64}=\frac{63}{64}\)
Đặt A = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
2A = \(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
2A - A = \(1-\frac{1}{64}\)
=> A = \(\frac{63}{64}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{32}-\frac{1}{64}\)
\(=1-\frac{1}{64}\)
=\(=\frac{63}{64}\)