1/2    + 1/4     + 1/8  + 1/16  + 1/32 + 1/64

 = 32/64 + 16/64 + 8/64 + 4/64 + 2/64 + 1/64

 =                         63/64

Chúc bạn học tốt nha!^-^

=63nha ban

         A =    \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\)+ \(\dfrac{1}{32}\)+\(\dfrac{1}{64}\)+\(\dfrac{1}{128}\)

A\(\times\) 2 =  1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\)+ \(\dfrac{1}{32}\)+ \(\dfrac{1}{64}\) 

A \(\times\) 2 - A = 1 - \(\dfrac{1}{128}\)

A\(\times\)(2-1) = \(\dfrac{128-1}{128}\)

A           = \(\dfrac{127}{128}\)

Gọi \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\) là B

\(B=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\)

\(2\cdot B=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{1}{32}+\dfrac{1}{64}\)

\(2\cdot B-B=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{1}{32}+\dfrac{1}{64}-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\right)\)

\(B=1+\left(\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+.....+\dfrac{1}{64}-\dfrac{1}{64}\right)-\dfrac{1}{128}\)

\(B=1+0-\dfrac{1}{128}\)

\(B=1-\dfrac{1}{128}\)

\(B=\dfrac{128}{128}-\dfrac{1}{128}\)

\(B=\dfrac{127}{128}\)

???

Are you sure?

nói chung là :

làm đc vế a là làm đc vế b

$A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}$ 

$=\frac{16}{32}+\frac{8}{32}+\frac{4}{32}+\frac{2}{32}+\frac{1}{32}$ 

$=\frac{31}{32}$ 

Xét: $1-A=1-\frac{31}{32}=\frac{1}{32}$ 

Xét: $1-\frac{2023}{2024}=\frac{1}{2024}$ 

Vì: $32<2024$ 

$\Rightarrow \frac{1}{32}>\frac{1}{2024}$ 

$\Rightarrow 1-\frac{1}{32}<1-\frac{1}{2024}$ 

$\Rightarrow \frac{31}{32}<\frac{2023}{2024}$ 

$Vậy:A<\frac{2023}{2024}$ 

\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+...+\dfrac{1}{x}=\dfrac{127}{256}\)

Đặt VT là A

\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{2}{x}\)

\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{2}{x}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{x}\right)=\dfrac{127}{256}\)

\(\Leftrightarrow A=1-\dfrac{1}{x}=\dfrac{127}{256}\)

\(\Leftrightarrow\dfrac{1}{x}=\dfrac{129}{256}\)

\(\Rightarrow x=\dfrac{256}{129}\)

\(\dfrac{7}{19}x\dfrac{8}{23}+\dfrac{7}{19}x\dfrac{15}{23}+1\dfrac{7}{19}\)

= \(\dfrac{7}{19}x\left(\dfrac{8}{23}+\dfrac{15}{23}\right)+1+\dfrac{7}{19}\)

=\(\dfrac{7}{19}x1+1+\dfrac{7}{19}\)

= \(\dfrac{7}{19}+1+\dfrac{7}{19}=1\dfrac{14}{19}\) = \(\dfrac{33}{19}\)

\(\dfrac{75}{100}+\dfrac{18}{21}+\dfrac{49}{32}+\dfrac{1}{4}+\dfrac{3}{21}-\dfrac{17}{32}\)

=  \(\dfrac{3}{4}+\dfrac{6}{7}+\dfrac{49}{32}+\dfrac{1}{4}+\dfrac{1}{7}-\dfrac{17}{32}\)

= \(\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\left(\dfrac{6}{7}+\dfrac{1}{7}\right)+\left(\dfrac{49}{32}-\dfrac{17}{32}\right)\)

= 1 + 1 + 1 = 3

\(\dfrac{8}{9}x\dfrac{15}{16}x\dfrac{24}{25}x\dfrac{35}{36}x\dfrac{48}{49}x\dfrac{63}{64}\)

= \(\dfrac{3}{4}\) *Câu này bạn tự sử dụng gạch nhé!

`1,`

`a,`

`7/19 \times 8/23 + 7/19 \times 15/23 + 1 7/19`

`= 7/19 \times 8/23 + 7/19 \times 15/23 + 1 + 7/19`

`= 7/19 \times (8/23 + 15/23 + 1) + 1`

`= 7/19 \times 2 + 1`

`=14/19 + 1`

`= 33/19`

`b,`

`75/100 + 18/21 + 49/32 + 1/4 + 3/21 - 17/32`

`= 75/100 + (18/21 + 3/21) + (49/32 - 17/32) + 1/4`

`= 0,75 + 1 + 1 + 0,25`

`= (0,75 + 0,25) + 1 + 1`

`= 1+1+1=3`

`c,`

`8/9 \times 15/16 \times 24/25 \times 35/36 \times 48/49 \times 63/64`

`=` \(\dfrac{2\times3}{3\times3}\times\dfrac{3\times5}{4\times4}\times\dfrac{3\times4\times2}{5\times5}\times\dfrac{5\times7}{6\times6}\times\dfrac{6\times8}{7\times7}\times\dfrac{7\times9}{8\times8}\)

`= 3/4` (bạn sử dụng gạch, rút gọn các số là được nhé).

a: \(A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^7\)

=>\(2\cdot A=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^6\)

=>\(2A-A=1-\left(\dfrac{1}{2}\right)^7=1-\dfrac{1}{128}=\dfrac{127}{128}\)

=>\(A=\dfrac{127}{128}\)

b: \(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{10\cdot11}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{10}-\dfrac{1}{11}\)

\(=1-\dfrac{1}{11}=\dfrac{10}{11}\)

ý a bạn nhé

`->D` 

vì : `4/16=(4:4)/(16:4)=1/4`

      `2/8=(2:2)/(8:2)=1/4`

      `3/12=(3:3)/(12:3)=1/4`

1+2+4+8+16+32+64+128+256+512+1024+2048

=1+(2+8)+(4+16)+(32+128)+(64+256)+(512+2048)+1024

=1+10+20+160+320+2560+1024

=4095

 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024 + 2048 = 4095 

k nha      công chúa nụ cười    =_=   ^_^

= 4095 bạn nhé ! k mk nha !

A =             1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\)+ \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\)+ \(\dfrac{1}{64}\)+ \(\dfrac{1}{128}\)

A\(\times\)2 = 2 + 1 + \(\dfrac{1}{2}\) +  \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\)

A \(\times\) 2 - A = 2 - \(\dfrac{1}{128}\)

A \(\times\)( 2-1) = \(\dfrac{255}{128}\)

A = \(\dfrac{255}{128}\)

Gọi \(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\) là T

\(T=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\)

\(2T=2+1+\dfrac{1}{2}+\dfrac{1}{4}+....+\dfrac{1}{64}\)

\(2T-T=\left(2+1+\dfrac{1}{2}+\dfrac{1}{4}+....+\dfrac{1}{64}\right)-\left(1+\dfrac{1}{2}+....+\dfrac{1}{64}+\dfrac{1}{128}\right)\)

\(T=2+\left(1-1\right)+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+....+\left(\dfrac{1}{64}-\dfrac{1}{64}\right)-\dfrac{1}{128}\)

\(T=2+0+0+...-\dfrac{1}{128}\)

\(T=\dfrac{256}{128}-\dfrac{1}{128}\)

\(T=\dfrac{255}{128}\)