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12 tháng 10 2017

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 = 2y , 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

12 tháng 10 2017

sau đăng vào box toán nhe bạn

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:

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

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

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

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

$\Rightarrow 3x=1023$

$\Rightarrow x=341$

27 tháng 2 2023

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

15 tháng 10 2021
Đáp án𝑠=15376
15 tháng 10 2021

Cả lời giải bn

\(\Leftrightarrow\dfrac{15}{16}:x=1-\dfrac{1}{12}=\dfrac{11}{12}\)

\(\Leftrightarrow x=\dfrac{15}{16}:\dfrac{11}{12}=\dfrac{15}{16}\cdot\dfrac{12}{11}=\dfrac{45}{44}\)

`#040911`

`a)`

`3 1/3 x + 16 3/4 = -13,25`

`=> 3 1/3 x = -13,25 - 16 3/4`

`=> 3 1/3 x = -30`

`=> x = -30 \div 3 1/3`

`=> x =-9`

Vậy, `x = -9`

`b)`

`3 2/7*x - 1/8 = 2 3/4`

`=> 3 2/7x = 2 3/4 + 1/8`

`=> 3 2/7x = 23/8`

`=> x = 23/8 \div 3 2/7`

`=> x = 7/8`

Vậy, `x = 7/8`

`c)`

`x \div 4 1/3 = -2,5`

`=> x = -2,5 * 4 1/3`

`=> x = -65/6`

Vậy, `x = -65/6`

`d)`

`( (3x)/7 + 1) \div (-4) = (-1)/28`

`=> (3x)/7 +1 = (-1)/28 * (-4)`

`=> (3x)/7 + 1 = 1/7`

`=> (3x)/7 = 1/7 - 1`

`=> (3x)/7 = -6/7`

`=> 3x = -6`

`=> x = -6 \div 3`

`=> x = -2`

Vậy, `x = -2.`

18 tháng 8 2023

a

=>10/3 . x + 16 + 3/4 = -13,25

=>10/3 x + 3/4 = -29,25

=>10/3 x = -30

=>x=-30 : 10/3

=>x=-30 . 3/10

=>x=-9

b.

=>23/7 x - 1/8 = = 11/4

=>23/7 x = 11/4 + 1/8

=>23/7 x= 22/8 + 1/8

=>23/7 x= 23/8

=>x=23/8 : 23/7

=>x=23/8 . 7/23

=>x=7/8

c.

=>x : 13/3 =-5/2

=>x=-5/2 . 13/3

=>x=-65/6

d.

=>3x/7 +1 = (-1/28) . (-4)

=>3x/7 + 1 = 1/7

=>3x/7 = -6/7

=>3x=-6

=>x=-2

3 tháng 12 2018

\(\dfrac{1}{x-1}-\dfrac{1}{x+1}-\dfrac{2}{x^2+1}-\dfrac{4}{x^4+1}-\dfrac{8}{x^8+1}-\dfrac{16}{x^{16}+1}\)

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

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

\(=\dfrac{2\left(x^2+1-x^2+1\right)}{x^4-1}-\dfrac{4}{x^4+1}-\dfrac{8}{x^8+1}-\dfrac{16}{x^{16}+1}\)

\(=\dfrac{4}{x^4-1}-\dfrac{4}{x^4+1}-\dfrac{8}{x^8+1}-\dfrac{16}{x^{16}+1}\)

\(=\dfrac{4\left(x^4+1-x^4+1\right)}{x^8-1}-\dfrac{8}{x^8+1}-\dfrac{16}{x^{16}+1}\)

\(=\dfrac{8}{x^8-1}-\dfrac{8}{x^8+1}-\dfrac{16}{x^{16}+1}\)

\(=\dfrac{8\left(x^8+1-x^8+1\right)}{x^{16}-1}-\dfrac{16}{x^{16}+1}\)

\(=\dfrac{16}{x^{16}-1}-\dfrac{16}{x^{16}+1}\)

\(=\dfrac{16\left(x^{16}+1-x^{16}+1\right)}{x^{32}-1}\)

\(=\dfrac{32}{x^{32}-1}\)

29 tháng 11 2018

\(A=\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(A=\left(\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}+\dfrac{1-x}{\left(1+x\right)\left(1-x\right)}\right)+...+\dfrac{16}{1+x^{16}}\)

\(A=\dfrac{1+x+1-x}{1-x^2}+\dfrac{2}{1+x^2}+...+\dfrac{16}{1+x^{16}}\)

\(A=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+...+\dfrac{16}{1+x^{16}}\)

Tiếp tục các bước như ở dòng 2 và 3 ta có :

\(A=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)

\(A=\dfrac{16\left(1+x^{16}\right)}{\left(1-x^{16}\right)\left(1+x^{16}\right)}+\dfrac{16\left(1-x^{16}\right)}{\left(1+x^{16}\right)\left(1-x^{16}\right)}\)

\(A=\dfrac{16+16x^{16}+16-16x^{16}}{1-x^{32}}\)

\(A=\dfrac{32}{1-x^{32}}\)

17 tháng 12 2017

\(A=\dfrac{2}{x^2+2x}+\dfrac{2}{x^2+6x+8}+\dfrac{2}{x^2+10x+24}+\dfrac{2}{x^2+14x+48}\)

\(A=\dfrac{2}{x\left(x+2\right)}+\dfrac{2}{\left(x+2\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}+\dfrac{2}{\left(x+6\right)\left(x+8\right)}\)

\(A=\dfrac{1}{x}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+8}\)

\(A=\dfrac{1}{x}-\dfrac{1}{x+8}=\dfrac{x+8}{x\left(x+8\right)}-\dfrac{x}{\left(x+8\right)}=\dfrac{8}{x\left(x+8\right)}\)

\(B=\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(B=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(B=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(B=\dfrac{8}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(B=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)

\(B=\dfrac{32}{1-x^{32}}\)