Lời giải:
$\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}$

$< \frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}$

$=\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+...+\frac{100-99}{99.100}$

$=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}$

$=\frac{1}{3}-\frac{1}{100}< \frac{1}{3}$

Bài 2

a) x/(-3) = 4/7

x = 4/7 . (-3)

x = -12/7

b) -12/(-15) = x/25

x = 4/5 . 25

x = 20

c) (12 - x)/27 = 5/4

12 - x = 5/4 . 27

12 - x = 135/4

x = 12 - 135/4

x = -87/4

d) 12/(-3 - x) = 4/7

-3 - x = 12 : 4/7

-3 - x = 21

x = -3 - 21

x = -24

Bài 1

a) 21/35 = 3/5 = 15/25

13/(-25) = -13/25

b) 2/5 = 24/60

3/(-12) = -15/60

5/6 = 50/60

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:

�=12⋅15+115⋅3+13⋅21+121⋅4+...+187⋅90A=2⋅151​+15⋅31​+3⋅211​+21⋅41​+...+87⋅901​

13�=132⋅15+1315⋅3+133⋅21+1321⋅4+...+1387⋅9013A=2⋅1513​+15⋅313​+3⋅2113​+21⋅413​+...+87⋅9013​

13�=12−115+115−13+13−121+121−14+...+187−19013A=21​−151​+151​−31​+31​−211​+211​−41​+...+871​−901​

13�=12−19013A=21​−901​

13�=224513A=4522​

�=2245x13=22585A=45x1322​=58522​

  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