`2x-15=-25`

`2x=-10`

`x=-5`

___________

`3/5<x/10<4/5`

`3/5=(3xx10)/(5xx10)=30/50`

`x/10=(5x)/(10xx5)=(5x)/50`

`4/5=(4xx10)/(5xx10)=40/50`

`=>30/50<(5x)/50<40/50`

`=>30<5x<40`

`=>x=7`

\(=\left(\dfrac{15}{2021}+\dfrac{16}{2022}-\dfrac{115}{2023}\right)\cdot\dfrac{3-2-1}{6}=0\)

\(\left(\dfrac{15}{2021}+\dfrac{16}{2022}-\dfrac{115}{2023}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)=\left(\dfrac{15}{2021}+\dfrac{8}{1011}-\dfrac{115}{2023}\right)\cdot\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)=\left(\dfrac{15}{2021}+\dfrac{8}{1011}-\dfrac{115}{2023}\right)\cdot0=0\)

...

\(A=\dfrac{2023^{2022+2}}{2023^{2022-1}}=2023^{2024-2021}=2023^3\\ B=\dfrac{2023^{2022}}{2023^{2022-3}}=2023^3\\ \Rightarrow A=B\left(=2023^3\right)\)

Haha

a, \(\dfrac{7}{22}\) - \(\dfrac{15}{23}\) + \(\dfrac{2022}{2023}\) - \(\dfrac{8}{23}\) + \(\dfrac{15}{22}\)

= ( \(\dfrac{7}{22}\) + \(\dfrac{15}{22}\)) - ( \(\dfrac{15}{23}+\dfrac{18}{23}\)) + \(\dfrac{2022}{2023}\)

= \(\dfrac{22}{22}\) - \(\dfrac{23}{23}\) + \(\dfrac{2022}{2023}\)

= 1 - 1 + \(\dfrac{2022}{2023}\)

= \(\dfrac{2022}{2023}\) 

b, - \(\dfrac{2}{11}\) + 5\(\dfrac{5}{6}\) ( 14\(\dfrac{1}{5}\) - 11\(\dfrac{1}{5}\)): 5\(\dfrac{1}{2}\)

= - \(\dfrac{2}{11}\) + \(\dfrac{35}{6}\) ( \(\dfrac{71}{5}\) - \(\dfrac{56}{5}\)) : \(\dfrac{11}{2}\)

= - \(\dfrac{2}{11}\) + \(\dfrac{35}{6}\) . \(\dfrac{15}{5}\) : \(\dfrac{11}{2}\)

= - \(\dfrac{2}{11}\) + \(\dfrac{35}{2}\) \(\times\) \(\dfrac{2}{11}\)

= - \(\dfrac{2}{11}\) + \(\dfrac{35}{11}\)

= \(\dfrac{33}{11}\)

= 3 

c, 2000 + { 20 - [ 4.2022⁰ - (3² + 5):2] }

= 2000 + { 20 - [ 4.1 - (9+5):2]}

= 2000 + { 20 - [ 4 - 14 : 2 ]}

= 2000 + { 20 - [ 4 -7]}

= 2000 + { 20 - (-3)}

= 2000 + 23

= 2023

a: \(\frac{7}{22}+\frac{-15}{23}+\frac{2022}{2023}+\frac{-8}{23}+\frac{15}{22}\)

\(=\left(\frac{7}{22}+\frac{15}{22}\right)+\left(-\frac{15}{23}-\frac{8}{23}\right)+\frac{2022}{2023}\)

\(=1-1+\frac{2022}{2023}=\frac{2022}{2023}\)

b: \(-\frac{2}{11}+5\frac56\left(14\frac15-11\frac15\right):5\frac12\)

\(=-\frac{2}{11}+\frac{35}{6}\left(14+\frac15-11-\frac15\right):\frac{11}{2}\)

\(=\frac{-2}{11}+\frac{35}{6}\cdot\frac{2}{11}\cdot3=\frac{-2}{11}+\frac{35}{11}=\frac{33}{11}=3\)

c: \(2000+\left\lbrace20-\left\lbrack4\cdot2022^0-\left(3^2+5\right):2\right\rbrack\right\rbrace\)

\(=2000+20-\left\lbrack4\cdot1-14:2\right\rbrack\)

=2020-(4-7)

=2020-(-3)

=2023

a:

Sửa đề: \(S=1-3+5-7+...+2021-2023+2025\)

Từ 1 đến 2025 sẽ có:

\(\dfrac{2025-1}{2}+1=\dfrac{2024}{2}+1=1013\left(số\right)\)

Ta có: 1-3=5-7=...=2021-2023=-2

=>Sẽ có \(\dfrac{1013-1}{2}=\dfrac{1012}{2}=506\) cặp có tổng là -2 trong dãy số này

=>\(S=506\cdot\left(-2\right)+2025=2025-1012=1013\)

b: \(S=1+2-3-4+5+6-7-8+...+2021+2022-2023-2024\)

Từ 1 đến 2024 là: \(\dfrac{\left(2024-1\right)}{1}+1=2024\left(số\right)\)

Ta có: 1+2-3-4=5+6-7-8=...=2021+2022-2023-2024=-4

=>Sẽ có \(\dfrac{2024}{4}=506\) cặp có tổng là -4 trong dãy số này

=>\(S=506\cdot\left(-4\right)=-2024\)

Sửa đề: \(\frac{x-1}{2023}+\frac{x-2}{2022}+\cdots+\frac{x-2022}{2}=2022\)

Ta có: \(\frac{x-1}{2023}+\frac{x-2}{2022}+\cdots+\frac{x-2022}{2}=2022\)

=>\(\left(\frac{x-1}{2023}-1\right)+\left(\frac{x-2}{2022}-1\right)+\cdots+\left(\frac{x-2022}{2}-1\right)=2022-2022=0\)

=>\(\frac{x-2024}{2023}+\frac{x-2024}{2022}+\cdots+\frac{x-2024}{2}=0\)

=>\(\left(x-2024\right)\left(\frac{1}{2023}+\frac{1}{2022}+\cdots+\frac12\right)=0\)

=>x-2024=0

=>x=2024