\(\left(2x-3\right)^{2022}=\left(2x-3\right)^{2021}\)

\(\left(2x-3\right)^{2021}\left(2x-3-1\right)=0\)

\(\left(2x-3\right)^{2021}\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\)

Ta có: 

\(\left(2x-3\right)^{2022}=\left(2x-3\right)^{2021}\)

\(\Rightarrow\left(2x-3\right)^{2022}-\left(2x-3\right)^{2021}=0\)

\(\Rightarrow\left(2x-3\right)^{2021}\left[\left(2x-3\right)-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\2x-3-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=3\\2x=4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{3}{2};2\right\}\).

\(Toru\)

^ cái dấu này có phải là mũ ko bạn

a)\(27.65+27.35+300=27.\left(65+35\right)+300\)

\(=27.100+300=2700+300=3000\)

b)\(3838:\left[\left(190-6.5^2\right):4+3\right]\)

\(=3838:\left[\left(190-6.25\right):4+3\right]\)

\(=3838:\left[\left(190-150\right):4+3\right]\)

\(=3838:\left[40:4+3\right]=3838:\left[10+3\right]\)

\(=3838:13=\dfrac{3838}{13}\)

c)\(2022-x=2021\)

\(x=2022-2021=1\)

d)\(26+14:\left(x-5\right)=33\)

\(14:\left(x-5\right)=33-26=7\)

\(x-5=14+7=2\)

\(x=2+5=7\)

e)đề hỏi làm j thế bạn

câu a chưa đủ đề em hấy

c, \(x\)(\(x\) - 2022) + 4.(2022 - \(x\)) = 0

       (\(x\) - 2022).(\(x\) - 4) = 0

         \(\left[{}\begin{matrix}x-2022=0\\x+4=0\end{matrix}\right.\)

          \(\left[{}\begin{matrix}x=2022\\x=4\end{matrix}\right.\)

b, (\(x\) - 1)(\(x^2\) + \(x\) + 1) - \(x\)(\(x\) - 2)(\(x\) + 2)  = 7

     \(x^3\) - 1 - \(x\).(\(x^2\) - 4) = 7

      \(x^3\) - 1 - \(x^3\) + 4\(x\)  = 7

        (\(x^3\) - \(x^3\)) - 1 + 4\(x\) = 7

                       - 1 + 4\(x\) = 7

                               4\(x\) = 7 + 1

                               4\(x\) = 8

                                  \(x\) = 8:4

                                   \(x\) = 2

\(3^{x+2}+1=28\)

<=> \(3^{x+2}=27\)

<=> \(3^{x+2}=3^3\)

<=> x+2 = 3

<=> x =1

=>x-1/3=0 và 1/4-y=0

=>x=1/3 và y=1/4

x=5x1/5=1

Lời giải:
$\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{n(n+1)}=\frac{2022}{2023}$

$\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{n(n+1)}=\frac{2022}{2023}$

$2[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{n(n+1)}]=\frac{2022}{2023}$

$2[\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{n(n+1)}]=\frac{2022}{2023}$
$2(\frac{1}{2}-\frac{1}{n+1})=\frac{2022}{2023}$

$1-\frac{2}{n+1}=1-\frac{1}{2023}$

$\Rightarrow \frac{2}{n+1}=\frac{1}{2023}$

$\Rightarrow n+1=2.2023=4046$

$\Rightarrow n=4045$

a. \(\dfrac{2021+2020.2022}{2021.2022-1}\)

\(\dfrac{2021.2022-2022+2021}{2021.2022-1}=\dfrac{2021.2022-1}{2021.2022-1}=1\)

\(b.\dfrac{2022+2021.2023}{2022.2023-1}=\dfrac{2021.2023-2023+2022}{2022.2023-1}\)

\(=\dfrac{2021.2023-1}{2022.2023-1}\)

thanks các cậu nhiều