Ta có:

\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\)

...

\(\dfrac{1}{2023^2}< \dfrac{1}{2022\cdot2023}\) 

\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2023^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2022\cdot2023}\)

\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2023^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\)

\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2023^2}< 1-\dfrac{1}{2023}< 1\)

\(13\cdot28-13\cdot12+16\cdot7\)

\(=13\left(28-12\right)+16\cdot7\)

\(=13\cdot16+16\cdot7=16\left(13+7\right)=16\cdot20=320\)

Gọi biểu thức trên là A, ta có:

$A=\genfrac{}{}{0ex}{}{1}{2\cdot 15}+\genfrac{}{}{0ex}{}{1}{15\cdot 3}+\genfrac{}{}{0ex}{}{1}{3\cdot 21}+\genfrac{}{}{0ex}{}{1}{21\cdot 4}+...+\genfrac{}{}{0ex}{}{1}{87\cdot 90}$

$13A=\genfrac{}{}{0ex}{}{13}{2\cdot 15}+\genfrac{}{}{0ex}{}{13}{15\cdot 3}+\genfrac{}{}{0ex}{}{13}{3\cdot 21}+\genfrac{}{}{0ex}{}{13}{21\cdot 4}+...+\genfrac{}{}{0ex}{}{13}{87\cdot 90}$

$13A=\genfrac{}{}{0ex}{}{1}{2}-\genfrac{}{}{0ex}{}{1}{15}+\genfrac{}{}{0ex}{}{1}{15}-\genfrac{}{}{0ex}{}{1}{3}+\genfrac{}{}{0ex}{}{1}{3}-\genfrac{}{}{0ex}{}{1}{21}+\genfrac{}{}{0ex}{}{1}{21}-\genfrac{}{}{0ex}{}{1}{4}+...+\genfrac{}{}{0ex}{}{1}{87}-\genfrac{}{}{0ex}{}{1}{90}$

$13A=\genfrac{}{}{0ex}{}{1}{2}-\genfrac{}{}{0ex}{}{1}{90}$

$13A=\genfrac{}{}{0ex}{}{22}{45}$

$A=\genfrac{}{}{0ex}{}{22}{45\text{x}13}=\genfrac{}{}{0ex}{}{22}{585}$

  A = \(\dfrac{1}{7}\) + \(\dfrac{1}{7^2}\) + \(\dfrac{1}{7^3}\) + ... + \(\dfrac{1}{7^{100}}\)

7A = 7 + \(\dfrac{1}{7}\) +  \(\dfrac{1}{7^2}\) + ....+ \(\dfrac{1}{7^{100}}\)

7A - A = (7 + \(\dfrac{1}{7}\) + \(\dfrac{1}{7^2}\) +... + \(\dfrac{1}{7^{99}}\)) - (\(\dfrac{1}{7}\) + \(\dfrac{1}{7^2}\) + \(\dfrac{1}{7^3}\) + ... + \(\dfrac{1}{7^{100}}\))

6A = 7 + \(\dfrac{1}{7}\) + \(\dfrac{1}{7^2}\) + ... + \(\dfrac{1}{7^{99}}\) - \(\dfrac{1}{7}\)  - \(\dfrac{1}{7^2}\) - \(\dfrac{1}{7^3}\) - ... - \(\dfrac{1}{7^{100}}\)

6A = (\(\dfrac{1}{7}\) - \(\dfrac{1}{7}\)) + (\(\dfrac{1}{7^2}\) - \(\dfrac{1}{7^2}\)) + (\(\dfrac{1}{7^3}\) - \(\dfrac{1}{7^3}\)) +...+(\(\dfrac{1}{7^{99}}\) - \(\dfrac{1}{7^{99}}\))+ (7 - \(\dfrac{1}{7^{100}}\))

6A = 0 + 0 + ... + 0 + 7 - \(\dfrac{1}{7^{100}}\)

6A = 7 - \(\dfrac{1}{7^{100}}\)

A = (7 - \(\dfrac{1}{7^{100}}\)) : 6

A = \(\dfrac{7}{6}\) - \(\dfrac{1}{6.7^{100}}\)

   G =   \(\dfrac{3}{5}\) + \(\dfrac{3}{5^4}\) + \(\dfrac{3}{5^7}\) + ... + \(\dfrac{3}{5^{100}}\)

5³G =  75 + \(\dfrac{3}{5}\) + \(\dfrac{3}{5^4}\) +... + \(\dfrac{3}{5^{99}}\)

125G - G = (75 + \(\dfrac{3}{5}\) + \(\dfrac{3}{5^4}\) + \(\dfrac{3}{5^7}\) + ... + \(\dfrac{3}{5^{99}}\))  - (\(\dfrac{3}{5}\) + \(\dfrac{3}{5^4}\)+\(\dfrac{3}{5^7}\)+...+\(\dfrac{3}{5^{100}}\))

124G =  75 + \(\dfrac{3}{5}\) + \(\dfrac{3}{5^4}\) + \(\dfrac{3}{5^7}\)+...+ \(\dfrac{3}{5^{99}}\) - \(\dfrac{3}{5}\) - \(\dfrac{3}{5^4}\) - \(\dfrac{3}{5^7}\) - ... - \(\dfrac{3}{5^{100}}\) 

124G = (75 - \(\dfrac{3}{5^{100}}\)) + (\(\dfrac{3}{5}\) - \(\dfrac{3}{5}\)) +(\(\dfrac{3}{5^4}\) - \(\dfrac{3}{5^4}\)) +...+ (\(\dfrac{3}{5^{99}}\) - \(\dfrac{3}{5^{99}}\)) 

124G = 75 - \(\dfrac{3}{5^{100}}\) + 0 + 0 + ... + 0

124G =  75 - \(\dfrac{3}{5^{100}}\)

   G = (75 - \(\dfrac{3}{5^{100}}\)): 124

   G = \(\dfrac{75}{124}\) - \(\dfrac{3}{124.5^{100}}\)

Lời giải:
\(2022A=\frac{2022^{2024}+2022}{2022^{2024}+1}=1+\frac{2021}{2022^{2024}+1}< 1+\frac{2021}{2022^{2023}+1}=\frac{2022^{2023}+2022}{2022^{2023}+1}=2022B\)

$\Rightarrow A< B$

Lời giải:
$x+(x+1)+(x+2)+....+(x+30)=1240$

$\underbrace{(x+x+x+...+x)}_{31}+(1+2+3+....+30)=1240$

$31\times x+30\times 31:2=1240$

$31\times x+465=1240$

$31\times x=775$

$x=775:31=25$

a:

ĐKXĐ: x<>3

 \(\dfrac{x-3}{2}=\dfrac{72}{x-3}\)

=>\(\left(x-3\right)^2=72\cdot2=144\)

=>\(\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\left(nhận\right)\\x=-9\left(nhận\right)\end{matrix}\right.\)
b: \(\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\cdot x=\dfrac{1}{49}+\dfrac{2}{48}+...+\dfrac{48}{2}+\dfrac{49}{1}\)

=>\(x\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+...+\left(\dfrac{48}{2}+1\right)+1\)

=>\(x\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)=\dfrac{50}{49}+\dfrac{50}{48}+...+\dfrac{50}{2}+\dfrac{50}{50}\)

=>\(x\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)=50\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)

=>x=50

24km

24 km

\(C=\dfrac{1}{99\cdot97}-\dfrac{1}{97\cdot95}-...-\dfrac{1}{5\cdot3}-\dfrac{1}{3\cdot1}\)

\(=\dfrac{1}{99\cdot97}-\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{95\cdot97}\right)\)

\(=\dfrac{1}{97\cdot99}-\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{95\cdot97}\right)\)

\(=\dfrac{1}{97\cdot99}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)

\(=\dfrac{1}{97\cdot99}-\dfrac{1}{2}\left(1-\dfrac{1}{97}\right)=\dfrac{1}{97\cdot99}-\dfrac{48}{97}\)

\(=\dfrac{1-48\cdot99}{97\cdot99}=\dfrac{-4751}{9603}\)