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Đặt \(A=\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\) có:
\(2A=\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{512}\)
\(\Rightarrow2A-A=\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{512}\right)-\left(\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\right)\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{1024}\)
\(\Rightarrow\dfrac{1}{2}-\left(\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\right)=\dfrac{1}{2}-\left(\dfrac{1}{2}-\dfrac{1}{1024}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{1024}=\dfrac{1}{1024}\)
Vậy...
Cách của Tuấn Anh Phan Nguyễn đây.
\(=\dfrac{1}{2}-\left[\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{512}+\dfrac{1}{1024}\right]\)
\(=\dfrac{1}{2}-\left[\left(\dfrac{1}{2}-\dfrac{1}{4}\right)+\left(\dfrac{1}{4}-\dfrac{1}{8}\right)+\left(\dfrac{1}{8}-\dfrac{1}{16}\right)+...+\left(\dfrac{1}{512}-\dfrac{1}{1024}\right)\right]\)\(=\dfrac{1}{2}-\left(\dfrac{1}{2}-\dfrac{1}{1024}\right)=\dfrac{1}{1024}.\)
Đặt :
\(H=-1-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-..........-\dfrac{1}{1024}\)
\(\Leftrightarrow H=-1-\left(\dfrac{1}{2}+\dfrac{1}{4}+...........+\dfrac{1}{1024}\right)\)
Đặt :
\(T=\dfrac{1}{2}+\dfrac{1}{4}+.......+\dfrac{1}{1024}\)
\(\Leftrightarrow T=\dfrac{1}{2}+\dfrac{1}{2^2}+..........+\dfrac{1}{2^{10}}\)
\(\Leftrightarrow2T=1+\dfrac{1}{2}+\dfrac{1}{2^2}+.........+\dfrac{1}{2^9}\)
\(\Leftrightarrow2T-T=\left(1+\dfrac{1}{2}+.....+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{10}}\right)\)
\(\Leftrightarrow T=1-\dfrac{1}{2^{10}}\)
\(\Leftrightarrow H=-1-\left(1-\dfrac{1}{2^{10}}\right)\)
\(\Leftrightarrow H=-1-1+\dfrac{1}{2^{10}}\)
\(\Leftrightarrow H=-2+\dfrac{1}{2^{10}}\)
Đặt \(A=-1-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-...-\dfrac{1}{1024}\)
\(A=-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\right)\)
Đặt \(B=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\)
\(2B=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}\)
\(2B-B=1-\dfrac{1}{1024}\)
\(\Rightarrow B=\dfrac{1023}{1024}\)
\(\Rightarrow A=-\dfrac{1023}{1024}\)
\(\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot5-\left(\dfrac{1}{4}\right)^5\cdot3}{\dfrac{1}{1024}\cdot\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^{11}}\)
\(=\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot2}{\left(\dfrac{1}{2}\right)^{10}\cdot\left(\dfrac{1}{3}-\dfrac{1}{2}\right)}\)
\(=2:\dfrac{-1}{6}=2\cdot\left(-6\right)=-12\)
\(a.\)
\(1-\dfrac{1}{2}\left(\dfrac{3}{2}-2x\right)=4x-\dfrac{1}{4}\)
\(\Rightarrow1-\dfrac{3}{4}+x=4x-\dfrac{1}{4}\)
\(\Rightarrow1-\dfrac{3}{4}+\dfrac{1}{4}=4x-x\)
\(\Rightarrow3x=\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{6}\)
\(b.\)
\(x^{10}=1024\)
\(\Rightarrow x^{10}=2^{10}\)
\(\Rightarrow x=2\)
\(c.\)
\(3^x=81\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
\(B=\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot5-\left(\dfrac{1}{2}\right)^{10}\cdot3}{\left(\dfrac{1}{2}\right)^{10}\cdot\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^{11}}\\ =\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot\left(5-3\right)}{\left(\dfrac{1}{2}\right)^{10}\cdot\left(\dfrac{1}{3}-\dfrac{1}{2}\right)}\\ =\dfrac{2}{-\dfrac{1}{6}}\\ =-12\)
\(B=\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot5-\left(\dfrac{1}{2}\right)^{10}\cdot3}{\left(\dfrac{1}{2}\right)^{10}\cdot\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^{11}}\\ B=\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot\left(5-3\right)}{\left(\dfrac{1}{2}\right)^{10}\cdot\left(\dfrac{1}{3}-\dfrac{1}{2}\right)}\\ B=\dfrac{2}{-\dfrac{1}{6}}\\ B=-12\)
Đặt \(B=1+\dfrac{1}{2}+...+\dfrac{1}{1024}\) và \(A=-1-\dfrac{1}{2}-\dfrac{1}{4}-...-\dfrac{1}{1024}\)
=>A=-B
\(B=1+\dfrac{1}{2}+...+\dfrac{1}{1024}\)
=>\(\dfrac{1}{2}B=\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2^{11}}\)
=>\(-\dfrac{1}{2}B=\dfrac{1}{2^{11}}-1\)
=>\(\dfrac{1}{2}B=1-\dfrac{1}{2^{11}}=\dfrac{2^{11}-1}{2^{11}}\)
=>\(B=\dfrac{2^{11}-1}{2^{10}}\)
=>\(A=\dfrac{1-2^{11}}{2^{10}}\)
\(\dfrac{3}{16}\) - (\(x\) - \(\dfrac{5}{4}\)) - ( \(\dfrac{3}{4}\) - \(\dfrac{7}{8}\) - 1) = 2\(\dfrac{1}{2}\)
\(\dfrac{3}{16}\) - \(x\) + \(\dfrac{5}{4}\) - \(\dfrac{3}{4}\) + \(\dfrac{7}{8}\) + 1 = \(\dfrac{5}{2}\)
\(\dfrac{3}{16}\) - \(x\) + ( \(\dfrac{5}{4}\) - \(\dfrac{3}{4}\)) + (\(\dfrac{7}{8}\) + 1) = \(\dfrac{5}{2}\)
\(\dfrac{3}{16}\) - \(x\) + \(\dfrac{1}{2}\) + \(\dfrac{15}{8}\) = \(\dfrac{5}{2}\)
( \(\dfrac{3}{16}\) + \(\dfrac{1}{2}\) + \(\dfrac{15}{8}\)) - \(x\) = \(\dfrac{5}{2}\)
\(\dfrac{41}{16}\) - \(x\) = \(\dfrac{5}{2}\)
\(x\) = \(\dfrac{41}{16}\) - \(\dfrac{5}{2}\)
\(x\) = \(\dfrac{1}{16}\)
2, \(\dfrac{1}{2}\).( \(\dfrac{1}{6}\) - \(\dfrac{9}{10}\)) = \(\dfrac{1}{5}\) - \(x\) + ( \(\dfrac{1}{15}\) - \(\dfrac{-1}{5}\))
\(\dfrac{1}{2}\).(-\(\dfrac{11}{15}\)) = \(\dfrac{1}{5}\) - \(x\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{5}\)
- \(\dfrac{11}{30}\) = ( \(\dfrac{1}{5}\)+ \(\dfrac{1}{5}\)+ \(\dfrac{1}{15}\)) - \(x\)
- \(\dfrac{11}{30}\) = \(\dfrac{7}{15}\) - \(x\)
\(x\) = \(\dfrac{7}{15}\) + \(\dfrac{11}{30}\)
\(x\) = \(\dfrac{5}{6}\)
`@` `\text {Ans}`
`\downarrow`
`1,`
`3/16 - (x - 5/4) - (3/4 + (-7)/8 - 1) = 2 1/2`
`=> 3/16 - x + 5/4 - (-1/8 - 1) = 2 1/2`
`=> 3/16 - x + 5/4 - (-9/8) = 2 1/2`
`=> 3/16 - x + 19/8 = 2 1/2`
`=> 3/16 - x = 2 1/2 - 19/8`
`=> 3/16 - x =1/8`
`=> x = 3/16 - 1/8`
`=> x = 1/16`
Vậy, `x = 1/16`
`2,`
`1/2* (1/6 - 9/10) = 1/5 - x + (1/15 - (-1)/5)`
`=> 1/2 * (-11/15) = 1/5 - x + 4/15`
`=> -11/30 = x + 1/5 - 4/15`
`=> x + (-1/15) = -11/30`
`=> x = -11/30 + 1/15`
`=> x = -3/10`
Vậy, `x = -3/10.`
\(\dfrac{x}{1024}=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...-\dfrac{1}{1024}\)
\(\dfrac{2x}{1024}=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+...-\dfrac{1}{512}\)
\(\Rightarrow\dfrac{x}{1024}+\dfrac{2x}{1024}=1-\dfrac{1}{1024}\)
\(\Rightarrow\dfrac{3x}{1024}=\dfrac{1023}{1024}\)
\(\Rightarrow3x=1023\)
\(\Rightarrow x=341\)
Lời giải:
$\frac{x}{1024}=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+...-\frac{1}{1024}$
$\frac{2x}{1024}=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+...-\frac{512}$
$\Rightarrow \frac{x}{1024}+\frac{2x}{1024}=1-\frac{1}{1024}$
$\frac{3x}{1024}=\frac{1023}{1024}$
$\Rightarrow 3x=1023$
$\Rightarrow x=341$