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\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+..........+\dfrac{1}{256}+\dfrac{1}{512}=?\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+...+\dfrac{1}{256}-\dfrac{1}{512}-\dfrac{1}{512}\)
\(=1-\dfrac{1}{512}\)
\(=\dfrac{511}{512}\)
Vậy \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+.........+\dfrac{1}{256}+\dfrac{1}{512}=\dfrac{511}{512}\)
Bài bạn trên cách trình bày mk ko hiểu lắm! mk làm lại nhé!
Đặt :
\(S=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...........+\dfrac{1}{256}+\dfrac{1}{512}\)
\(\Leftrightarrow S=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.........+\dfrac{1}{2^8}+\dfrac{1}{2^9}\)
\(\Leftrightarrow2S=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+.........+\dfrac{1}{2^9}\right)\)
\(\Leftrightarrow2S=1+\dfrac{1}{2}+\dfrac{1}{2^2}+.........+\dfrac{1}{2^8}\)
\(\Leftrightarrow2S-S=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^8}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+......+\dfrac{1}{2^9}\right)\)
\(\Leftrightarrow S=1-\dfrac{1}{2^9}\)
\(\Leftrightarrow S=1-\dfrac{1}{512}=\dfrac{511}{512}\)
\(x:\dfrac{1}{2}+x:\dfrac{1}{4}+x:\dfrac{1}{8}+...+x:\dfrac{1}{512}=511\\ 2x+4x+8x+..+512x=511\\ x\left(2+4+8+...+512\right)=511\\ x\left(2^1+2^2+2^3+...+2^9\right)=511\\ \)
Gọi \(S=2^1+2^2+2^3+...+2^9\)
\(2S=2^2+2^3+2^4+...+2^{10}\\ 2S-S=\left(2^2+2^3+2^4+...+2^{10}\right)-\left(2^1+2^2+2^3+...+2^9\right)\\ S=2^{10}-2\)
\(x\left(2^{10}-2\right)=511\\ 2x\left(2^9-1\right)=511\\ 2x\left(512-1\right)=511\\ 2x\cdot511=511\\ 2x=1\\ x=\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{2}\)
a) \(1-\dfrac{1}{2}=\dfrac{1}{2}\)
\(\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{3-2}{6}=\dfrac{1}{6}\)
\(\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{4-3}{12}=\dfrac{1}{12}\)
\(\dfrac{1}{4}-\dfrac{1}{5}=\dfrac{5-4}{20}=\dfrac{1}{20}\)
\(\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{6-5}{30}=\dfrac{1}{30}\)
b) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)
\(=\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+\left(\dfrac{1}{4}-\dfrac{1}{5}\right)+\left(\dfrac{1}{5}+\dfrac{1}{6}\right)\)
\(=1+\left(-\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+\left(-\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(-\dfrac{1}{5}+\dfrac{1}{5}\right)+-\dfrac{1}{6}\)\(=1+-\dfrac{1}{6}\)
\(=\dfrac{5}{6}\)
\(E=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{8}+\dfrac{1}{2}+\dfrac{1}{12}\)
\(E=\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{1}{24}\right)\)
\(E=\dfrac{2}{2}+\dfrac{3}{6}+\left(\dfrac{1}{8}+\dfrac{3}{24}\right)\)
\(E=1+\dfrac{1}{2}+\left(\dfrac{1}{8}+\dfrac{1}{8}\right)\)
\(E=\left(\dfrac{2}{2}+\dfrac{1}{2}\right)+\dfrac{2}{8}\)
\(E=\dfrac{3}{2}+\dfrac{1}{4}\)
\(E=\dfrac{6}{4}+\dfrac{1}{4}\)
\(E=\dfrac{7}{4}\)
Bài 1 :
\(=\dfrac{2}{11}+\dfrac{4}{11}-\dfrac{6}{11}-\dfrac{3}{8}-\dfrac{5}{8}=0-1=-1\)
Bài 2 :
\(\Rightarrow3+x=8\Leftrightarrow x=5\)
Bài 3 :
\(\Leftrightarrow x-\dfrac{5}{11}=\dfrac{5}{4}\Leftrightarrow x=\dfrac{35}{44}\)
Bài 4 :
Trong 2 ngày An đọc được số quyên phần quyên sách
\(\dfrac{1}{11}+\dfrac{8}{11}=\dfrac{9}{11}\)( quyển sách )
đs : 9/11 quyển sách
Đặt \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{256}+\dfrac{1}{512}\)
\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{128}+\dfrac{1}{256}\)
\(\Rightarrow A=2A-A=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{128}+\dfrac{1}{256}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-...-\dfrac{1}{256}-\dfrac{1}{512}\)
\(\Rightarrow A=1-\dfrac{1}{512}=\dfrac{511}{512}\)
Đặt A=12+14+18+...+1256+1512A=12+14+18+...+1256+1512
⇒2A=1+12+14+...+1128+1256⇒2A=1+12+14+...+1128+1256
⇒A=2A−A=1+12+14+...+1128+1256−12−14−18−...−1256−1512⇒A=2A−A=1+12+14+...+1128+1256−12−14−18−...−1256−1512
⇒A=1−1512=511512