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

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

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

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

\(\Rightarrow3x=1023\)

\(\Rightarrow x=341\)

Lời giải:

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

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

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

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

$\Rightarrow 3x=1023$

$\Rightarrow x=341$

S=1/2+1/4+1/8+...+1/1024=(1/2)^1+(1/2)^2+(1/2)^3+...+(1/2)^10

2S=1+(1/2)^1+(1/2)^2+...+(1/2)^9

2S-S=1-(1/2)^10

vậy S=1-(1/2)^10

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

\(=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)

\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)

\(\Rightarrow2A-A=A=1-\dfrac{1}{2^{10}}\)

đây ko phải lớp 5 đúng ko ?

Đặt A=1/2+1/4+1/8+..+1/1024

Ax2=1+1/2+1/4+1/8+..+1/512( Nhân cả 2 vế với 2)

Ax2-A=(1+1/2+1/4+1/8+..+1/512)-(1/2+1/4+1/8+..+1/1024)

<=>A=1-1/1024

<=>A=1023/1024

Vậy biểu thức đã cho = 1023/1024

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

\(W=-1-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-....-\dfrac{1}{1024}\)

\(W=-1-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+....+\dfrac{1}{1024}\right)\)

\(W=-1-\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{10}}\right)\)

Đặt:

\(A=\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{10}}\)

\(2A=2\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{10}}\right)\)

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

\(2A-A=\left(1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{10}}\right)\)

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

Thay \(A\) vào \(W\) ta có:

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

Gọi \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{x}=\dfrac{1023}{1024}\)

\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{x}=\dfrac{1023}{1024}\)

VẬy x là một lũy thừa của 2. Đặt x = 2^y , ta có:
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^y}\)

\(\Rightarrow2A=1+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{y-1}}\)

\(\Rightarrow2A-A=1+\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{y-1}}-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^8}\right)\)

\(=A-\dfrac{1}{2^y}\)

Vậy \(1-\dfrac{1}{2^y}=\dfrac{1023}{1024}\Leftrightarrow\dfrac{1}{2^y}=\dfrac{1}{1024}\Leftrightarrow2^y=1024\Rightarrow x=1024\)

Vậy x = 1024

sau đăng vào box toán nhe bạn

Giải:

a) \(F=\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{240}\)

\(\Leftrightarrow F=\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{15.16}\)

\(\Leftrightarrow F=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{15}-\dfrac{1}{16}\)

\(\Leftrightarrow F=\dfrac{1}{2}-\dfrac{1}{16}\)

\(\Leftrightarrow F=\dfrac{7}{16}\)

Vậy ...

b) \(G=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)

\(\Leftrightarrow G=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+\dfrac{1}{3^5}\)

\(\Leftrightarrow\dfrac{1}{3}G=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+\dfrac{1}{3^5}+\dfrac{1}{3^6}\)

\(\Leftrightarrow\dfrac{2}{3}G=\dfrac{1}{3}-\dfrac{1}{3^6}\)

\(\Leftrightarrow G=\dfrac{\dfrac{1}{3}-\dfrac{1}{3^6}}{\dfrac{2}{3}}\)

\(\Leftrightarrow G=\dfrac{\left(\dfrac{1}{3}-\dfrac{1}{3^6}\right)3}{2}\)

\(\Leftrightarrow G=\dfrac{1-\dfrac{1}{3^5}}{2}\)

\(\Leftrightarrow G=\dfrac{\dfrac{3^5-1}{3^5}}{2}\)

\(\Leftrightarrow G=\dfrac{3^5-1}{3^5.2}\)

Vậy ...

c) Tương tự b)

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