x-(1/1.2 + 1/2.3 + 1/3.4 + ...+ 1/2022.2023)= -2024/2023

x-(1-1/2 + 1/2-1/3 + 1/3-1/4 + ... + 1/2022-1/2023)=-2024/2023

x-(1-1/2023)=-2024/2023

x-2022/2023=-2024/2023

x = -2024/2023+2022/2023

x = -2/2023

Vậy x = -2/2023

:(((

a

ĐK: \(x\ne5\)

\(\dfrac{x-5}{3}=\dfrac{-12}{5-x}\\ \Leftrightarrow\dfrac{x-5}{3}=\dfrac{12}{x-5}\\ \Leftrightarrow\left(x-5\right)^2=12.3=36\\ \Leftrightarrow\left\{{}\begin{matrix}x-5=6\\x-5=-6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=11\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

b

ĐK: \(x\ne0;x\ne-1\)

\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+....+\dfrac{2}{x\left(x+1\right)}=\dfrac{2023}{2024}\)

\(\Leftrightarrow\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+....+\dfrac{2}{x\left(x+1\right)}=\dfrac{2023}{2024}\\ \Leftrightarrow2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{x}.\dfrac{1}{x+1}\right)=\dfrac{2023}{2024}\\ \Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{2023}{2024}\\ \Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{2023}{4048}\\ \Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{2023}{4048}=\dfrac{1}{4048}\\ \Leftrightarrow4048=x+1\\ \Leftrightarrow x=4047\left(tm\right)\)

a: =>(x-5)/3=12/(x-5)

=>(x-5)^2=36

=>x-5=6 hoặc x-5=-6

=>x=11 hoặc x=-1

b: =>\(2\left(\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2023}{2024}\)

=>1/2-1/3+1/3-1/4+...+1/x-1/x+1=2023/4048

=>1/2-1/x+1=2023/4048

=>1/(x+1)=1/4048

=>x+1=4048

=>x=4047

Có phải đề như này ko ?

`7/1^2`.`2024/2023-7/2023`.`1/2`

`#``\text{Lócc}`

`7/1.2 . 2024/2023 - 7/2023 . 1/2`

`= 7/2 . 2024/2023 - 7/2023 . 1/2`

`= 7/1 . 1/2 . 2024/2023 - 7/2023 . 1/2`

`= 7 . 1/2. (2024/2023 - 7/2023 )`

`= 7. 1/2 .2017/2023`

`= 7/2 . 2017/2023`

`= 14189/4046`

(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = (\(\dfrac{2021}{2}+1\))+(\(\dfrac{2020}{3}+1\))+....+(\(\dfrac{1}{2022}+1\))

(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = \(\dfrac{2023}{2}\)+\(\dfrac{2023}{3}\)+....+ \(\dfrac{2023}{2022}\)

(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = 2023.( \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\))

vậy x= 2023

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

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

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

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

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

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

\(\Rightarrow3x=1023\)

\(\Rightarrow x=341\)

Lời giải:

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

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

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

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

$\Rightarrow 3x=1023$

$\Rightarrow x=341$

\(x=\sqrt{\dfrac{2\sqrt{3}+2-6\sqrt{3}}{2\sqrt{3}\left(2\sqrt{3}+2\right)}}=\sqrt{\dfrac{2-4\sqrt{3}}{2\sqrt{3}\left(2\sqrt{3}+2\right)}}\) ko tồn tại vì 2-4căn 3<0

\(\left(\dfrac{1}{3}\right)^2-\left(\dfrac{1}{9}-\dfrac{2023}{2024}\right)\)

\(=\dfrac{1}{9}-\dfrac{1}{9}+\dfrac{2023}{2024}\)

\(=\dfrac{2023}{2024}\)