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:

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

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

...

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

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

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

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

1 A

2 A

3 B

4 C

5 A

6 A

7 D

8 A

9 D

10 A

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

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

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

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