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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 \(B=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\)
\(2B=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}\)
\(2B-B=1-\dfrac{1}{1024}\)
\(\Rightarrow B=\dfrac{1023}{1024}\)
\(\Rightarrow A=-\dfrac{1023}{1024}\)
Đặt \(B=1+\dfrac{1}{2}+...+\dfrac{1}{1024}\) và \(A=-1-\dfrac{1}{2}-\dfrac{1}{4}-...-\dfrac{1}{1024}\)
=>A=-B
\(B=1+\dfrac{1}{2}+...+\dfrac{1}{1024}\)
=>\(\dfrac{1}{2}B=\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2^{11}}\)
=>\(-\dfrac{1}{2}B=\dfrac{1}{2^{11}}-1\)
=>\(\dfrac{1}{2}B=1-\dfrac{1}{2^{11}}=\dfrac{2^{11}-1}{2^{11}}\)
=>\(B=\dfrac{2^{11}-1}{2^{10}}\)
=>\(A=\dfrac{1-2^{11}}{2^{10}}\)
\(a.\)
\(1-\dfrac{1}{2}\left(\dfrac{3}{2}-2x\right)=4x-\dfrac{1}{4}\)
\(\Rightarrow1-\dfrac{3}{4}+x=4x-\dfrac{1}{4}\)
\(\Rightarrow1-\dfrac{3}{4}+\dfrac{1}{4}=4x-x\)
\(\Rightarrow3x=\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{6}\)
\(b.\)
\(x^{10}=1024\)
\(\Rightarrow x^{10}=2^{10}\)
\(\Rightarrow x=2\)
\(c.\)
\(3^x=81\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
1: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^6\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{18}\)
=>4x=18
hay x=9/2
2: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^{36}\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{108}\)
=>4x=108
hay x=27
3: \(\left(\dfrac{1}{81}\right)^x=\left(\dfrac{1}{27}\right)^4\)
\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{4x}=\left(\dfrac{1}{3}\right)^{12}\)
=>4x=12
hay x=3
\(B=\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot5-\left(\dfrac{1}{2}\right)^{10}\cdot3}{\left(\dfrac{1}{2}\right)^{10}\cdot\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^{11}}\\ =\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot\left(5-3\right)}{\left(\dfrac{1}{2}\right)^{10}\cdot\left(\dfrac{1}{3}-\dfrac{1}{2}\right)}\\ =\dfrac{2}{-\dfrac{1}{6}}\\ =-12\)
\(B=\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot5-\left(\dfrac{1}{2}\right)^{10}\cdot3}{\left(\dfrac{1}{2}\right)^{10}\cdot\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^{11}}\\ B=\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot\left(5-3\right)}{\left(\dfrac{1}{2}\right)^{10}\cdot\left(\dfrac{1}{3}-\dfrac{1}{2}\right)}\\ B=\dfrac{2}{-\dfrac{1}{6}}\\ B=-12\)
1: =>1/3:x=3/5-2/3=9/15-10/15=-1/15
=>x=-1/3:1/15=5
2: \(\Leftrightarrow x\cdot\dfrac{2}{3}-3=\dfrac{2}{5}\cdot\left(-10\right)=-4\)
=>x*2/3=-1
=>x=-3/2
3: \(\Leftrightarrow\dfrac{8}{3}:x=\dfrac{25}{12}:\dfrac{-3}{50}=\dfrac{25}{12}\cdot\dfrac{-50}{3}\)
hay x=-48/625
9: =>x=-2*3/1,5=-4
8: =>2/3:x=5/2:-3/10=5/2*(-10)/3=-50/6=-25/3
=>x=-2/3:25/3=-2/3*3/25=-2/25
\(\dfrac{x}{1024}=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...-\dfrac{1}{1024}\)
\(\dfrac{2x}{1024}=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+...-\dfrac{1}{512}\)
\(\Rightarrow\dfrac{x}{1024}+\dfrac{2x}{1024}=1-\dfrac{1}{1024}\)
\(\Rightarrow\dfrac{3x}{1024}=\dfrac{1023}{1024}\)
\(\Rightarrow3x=1023\)
\(\Rightarrow x=341\)
Lời giải:
$\frac{x}{1024}=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+...-\frac{1}{1024}$
$\frac{2x}{1024}=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+...-\frac{512}$
$\Rightarrow \frac{x}{1024}+\frac{2x}{1024}=1-\frac{1}{1024}$
$\frac{3x}{1024}=\frac{1023}{1024}$
$\Rightarrow 3x=1023$
$\Rightarrow x=341$