\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2024}}\\ =>2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2023}}\\ =>2A-A=A=1-\dfrac{1}{2^{2024}}=\dfrac{2^{2024}-1}{2^{2024}}\)

e cảm ơn

\(\dfrac{x}{-2}=-\dfrac{8}{x}\left(x\ne0\right)\\ =>x.x=\left(-2\right).\left(-8\right)\\ =>x^2=16\\ =>x=\pm4\)

\(\dfrac{x}{-2}=\dfrac{-8}{x}\\ x.x=\left(-8\right).\left(-2\right)=16=4.4=\left(-4\right).\left(-4\right)\\ Vậy:\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

\(\dfrac{x+3}{2015}+\dfrac{x+2}{2016}+\dfrac{x+1}{2017}+\dfrac{x}{1009}=-5\\ =>\left(\dfrac{x+3}{2015}+1\right)+\left(\dfrac{x+2}{2016}+1\right)+\left(\dfrac{x+1}{2017}+1\right)+\left(\dfrac{x}{1009}+2\right)=0\\ =>\dfrac{x+2018}{2015}+\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}+\dfrac{x+2018}{1009}=0\\ =>\left(x+2018\right)\left(\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{1009}\right)=0\\ =>x+2018=0\\ =>x=-2018\left(TM\right)\)

em cảm ơn

Bài còn lại mờ quá em ơi

Lời giải:

$2^{x+2}.3^2-3.5^2=69$

$2^{x+3}.9=69+3.5^2=144$

$2^{x+3}=144:9=16=2^4$

$\Rightarrow x+3=4$

$\Rightarrow x=1$

\(2^{x+2}.3^2-3.5^2=69\\ =>2^{x+2}.9-3.25=69\\ =>2^{x+2}.9=69+75=144\\ =>2^{x+2}=144:9\\ =>2^{x+2}=16=2^4\\ =>x+2=4\\ =>x=2\)

\(\left(-4\right)x=36\)

\(x=36:\left(-4\right)\)

\(x=-9\)

$\left(-4\right)x=36$

$x=36:\left(-4\right)$

$x=-9$

Lời giải:

$\frac{1}{50}> \frac{1}{100}$

$\frac{1}{51}> \frac{1}{100}$

.....

$\frac{1}{98}> \frac{1}{100}$

$\frac{1}{99}> \frac{1}{100}$

$\Rightarrow S> \underbrace{\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}}_{50}=\frac{1}{100}.50=\frac{1}{2}$

vd 2 và 9 nha =)

a; 2^\(x\) : 16 = 2²⁰¹⁶

    2\(^x\)        = 2²⁰¹⁶ x 16

   2\(^x\)        = 2²⁰¹⁶ x 2⁴

   2\(^x\)        = 2²⁰²⁰

    \(x\)        = 2020

b; 3\(^{x+1}\) - 2 = 3² + [5² - 3.(2² - 1]

    3\(^{x+1}\) - 2 = 9 + [25 - 3.3]

    3\(^{x+1}\) - 2 = 9 + 16

   3\(^{x+1}\) - 2  = 25

  3\(^{x+1}\)        = 25 + 7

  3\(^{x+1}\)       = 27

3\(^{x+1}\)        = 3³

   \(x\) + 1 = 3

    \(x\)      = 3  - 1

    \(x\)      = 2