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Đặt \(B=1+\dfrac{1}{2}+...+\dfrac{1}{1024}\) và \(A=-1-\dfrac{1}{2}-\dfrac{1}{4}-...-\dfrac{1}{1024}\)

=>A=-B

\(B=1+\dfrac{1}{2}+...+\dfrac{1}{1024}\)

=>\(\dfrac{1}{2}B=\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2^{11}}\)

=>\(-\dfrac{1}{2}B=\dfrac{1}{2^{11}}-1\)

=>\(\dfrac{1}{2}B=1-\dfrac{1}{2^{11}}=\dfrac{2^{11}-1}{2^{11}}\)

=>\(B=\dfrac{2^{11}-1}{2^{10}}\)

=>\(A=\dfrac{1-2^{11}}{2^{10}}\)

31 tháng 12 2023

\(\dfrac{x}{1024}=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...-\dfrac{1}{1024}\)

\(\dfrac{2x}{1024}=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+...-\dfrac{1}{512}\)

\(\Rightarrow\dfrac{x}{1024}+\dfrac{2x}{1024}=1-\dfrac{1}{1024}\)

\(\Rightarrow\dfrac{3x}{1024}=\dfrac{1023}{1024}\)

\(\Rightarrow3x=1023\)

\(\Rightarrow x=341\)

AH
Akai Haruma
Giáo viên
30 tháng 12 2023

Lời giải:

$\frac{x}{1024}=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+...-\frac{1}{1024}$

$\frac{2x}{1024}=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+...-\frac{512}$

$\Rightarrow \frac{x}{1024}+\frac{2x}{1024}=1-\frac{1}{1024}$

$\frac{3x}{1024}=\frac{1023}{1024}$

$\Rightarrow 3x=1023$

$\Rightarrow x=341$

\(\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot5-\left(\dfrac{1}{4}\right)^5\cdot3}{\dfrac{1}{1024}\cdot\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^{11}}\)

\(=\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot2}{\left(\dfrac{1}{2}\right)^{10}\cdot\left(\dfrac{1}{3}-\dfrac{1}{2}\right)}\)

\(=2:\dfrac{-1}{6}=2\cdot\left(-6\right)=-12\)

6 tháng 9 2017

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 :

\(T=\dfrac{1}{2}+\dfrac{1}{4}+.......+\dfrac{1}{1024}\)

\(\Leftrightarrow T=\dfrac{1}{2}+\dfrac{1}{2^2}+..........+\dfrac{1}{2^{10}}\)

\(\Leftrightarrow2T=1+\dfrac{1}{2}+\dfrac{1}{2^2}+.........+\dfrac{1}{2^9}\)

\(\Leftrightarrow2T-T=\left(1+\dfrac{1}{2}+.....+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{10}}\right)\)

\(\Leftrightarrow T=1-\dfrac{1}{2^{10}}\)

\(\Leftrightarrow H=-1-\left(1-\dfrac{1}{2^{10}}\right)\)

\(\Leftrightarrow H=-1-1+\dfrac{1}{2^{10}}\)

\(\Leftrightarrow H=-2+\dfrac{1}{2^{10}}\)

6 tháng 9 2017

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 \(B=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\)

\(2B=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}\)

\(2B-B=1-\dfrac{1}{1024}\)

\(\Rightarrow B=\dfrac{1023}{1024}\)

\(\Rightarrow A=-\dfrac{1023}{1024}\)

7 tháng 6 2017

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...

7 tháng 6 2017

Cách của Tuấn Anh Phan Nguyễn đây.

\(=\dfrac{1}{2}-\left[\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{512}+\dfrac{1}{1024}\right]\)

\(=\dfrac{1}{2}-\left[\left(\dfrac{1}{2}-\dfrac{1}{4}\right)+\left(\dfrac{1}{4}-\dfrac{1}{8}\right)+\left(\dfrac{1}{8}-\dfrac{1}{16}\right)+...+\left(\dfrac{1}{512}-\dfrac{1}{1024}\right)\right]\)\(=\dfrac{1}{2}-\left(\dfrac{1}{2}-\dfrac{1}{1024}\right)=\dfrac{1}{1024}.\)

31 tháng 1 2019

\(B=\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot5-\left(\dfrac{1}{2}\right)^{10}\cdot3}{\left(\dfrac{1}{2}\right)^{10}\cdot\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^{11}}\\ =\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot\left(5-3\right)}{\left(\dfrac{1}{2}\right)^{10}\cdot\left(\dfrac{1}{3}-\dfrac{1}{2}\right)}\\ =\dfrac{2}{-\dfrac{1}{6}}\\ =-12\)

31 tháng 1 2019

\(B=\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot5-\left(\dfrac{1}{2}\right)^{10}\cdot3}{\left(\dfrac{1}{2}\right)^{10}\cdot\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^{11}}\\ B=\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot\left(5-3\right)}{\left(\dfrac{1}{2}\right)^{10}\cdot\left(\dfrac{1}{3}-\dfrac{1}{2}\right)}\\ B=\dfrac{2}{-\dfrac{1}{6}}\\ B=-12\)

15 tháng 10 2018

cảm ơn bạn rất nhiều

24 tháng 9 2017

\(a.\)

\(1-\dfrac{1}{2}\left(\dfrac{3}{2}-2x\right)=4x-\dfrac{1}{4}\)

\(\Rightarrow1-\dfrac{3}{4}+x=4x-\dfrac{1}{4}\)

\(\Rightarrow1-\dfrac{3}{4}+\dfrac{1}{4}=4x-x\)

\(\Rightarrow3x=\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{1}{6}\)

\(b.\)

\(x^{10}=1024\)

\(\Rightarrow x^{10}=2^{10}\)

\(\Rightarrow x=2\)

\(c.\)

\(3^x=81\)

\(\Rightarrow3^x=3^4\)

\(\Rightarrow x=4\)

28 tháng 6 2017

a

= { 1*( 1+1/2+1/3+1/4) } / { 1 * ( 1-1/2 +1/3-1/4)} : { 3*(1+1/2+1/3+1/4)} / { 2*( 1-1/2 +1/3-1/4)}

Sau đó bn tự tính ra nhé cứ tính nhu bình thường sẽ ra.

Mà mình thấy máy câu này yêu cầu tính chứ có bảo tính theo cách hợp lí đâu? Vì thế bn cứ lấy máy tính tính như bình thường là được .

20 tháng 7 2017

Kết quả là : C1=\(\dfrac{2}{3}\)