Tính nhanh Phân số
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...............\frac{1}{512}\)
Giúp mình nha
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\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+....+\frac{1}{256}-\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{512}\)
\(=\frac{255}{512}\)
Vậy \(A=\frac{255}{512}\)
A=14 +18 +116 +132 +164 +1128 +1256 +1512
=12 −14 +14 −18 +....+1256 −1512
=12 −1512
=255512
Vậy A=255512
Phạm Long Khánh
Ta có : \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)
Đặ A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)(1)
=> 2A = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}+\frac{1}{2^9}\)(2)
Lấy (2) trừ (1) theo vế ta có :
2A - A = \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\right)\)
=> A = \(1-\frac{1}{2^{10}}=\frac{2^{10}-1}{2^{20}}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{10}}\)
\(\Leftrightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^9}\)
\(\Rightarrow2A-A=1-\frac{1}{2^{10}}=\frac{1023}{1024}\)
ta có: 1+1/2+2+1/4+...+9+1/512
=(1+2+3+4+...+9)+(1/2+1/4+...+1/512)
=45+(1/2+1/4+...+1/512)
gọi số hạng (1/2+1/4+...+1/512) là a ta được :
a=1/2+1/4+...+1/512
2a=1+1/2+1/4+1/8+...+1/256
2a-a=(1+1/2+1/4+...+1/256)-(1/2+1/4+...+1/512)
=1-1/512
=511/512
vậy kết quả của biểu thức đó là45+511/512
\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\)
=\(\frac{3}{1.2}+\frac{3}{2.4}+\frac{3}{4.8}+\frac{3}{8.16}+\frac{3}{16.32}\)
=\(\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{4}+\frac{3}{4}-\frac{3}{8}+\frac{3}{8}-\frac{3}{16}+\frac{3}{16}-\frac{3}{36}\)
=\(\frac{3}{1}-\frac{3}{36}\)=\(\frac{35}{12}\)
Ta có: \(\frac{1}{2}=1-\frac{1}{2}\); \(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\); \(\frac{1}{8}=\frac{1}{4}-\frac{1}{8}\); ...; \(\frac{1}{512}=\frac{1}{256}-\frac{1}{512}\); \(\frac{1}{1024}=\frac{1}{512}-\frac{1}{1024}\)
Vậy \(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{256}-\frac{1}{512}+\frac{1}{512}-\frac{1}{1024}\)
\(=1+1-\frac{1}{1024}\)
\(=2-\frac{1}{1024}=\frac{2047}{1024}\)
\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-....+\frac{1}{256}-\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{512}\)
\(=\frac{255}{512}\)
Vậy \(A=\frac{255}{512}\)
=1/2-1/4+1/4-1/8+1/8-....+1/156-1/152
=1/2-1/152
=255/512
A=255/512
= \(\frac{17}{8}:\frac{25}{14}-\left(15-\frac{40}{3}\right):\frac{25}{6}\)
= \(\frac{17}{8}.\frac{14}{25}-\left(\frac{45}{3}-\frac{40}{3}\right).\frac{6}{25}\)
= \(\frac{119}{100}-\frac{5}{3}.\frac{6}{25}\) = \(\frac{119}{100}-\frac{2}{5}\)
= \(\frac{119}{100}-\frac{40}{100}=\frac{79}{100}\)
Chúc bạn Hk tốt!!!!!
Đặt phân số trên là A,ta có:
A=1/2+1/4+1/8+...+1/512
=>2A=1+1/2+1/4+...+1/256
=>A=2A-A=(1+1/2+1/4+...+1/256)-(1/2+1/4+1/8+...+1/512)
=>A=1+1/2+1/4+...+1/256-1/2-1/4-1/8-...-1/512
=>A=1-1/512
=>A=512/512-1/512
=>A=511/512
_Hok tốt_