\(\dfrac{3x+5}{x-1}=\dfrac{3x-3+8}{x-1}=\dfrac{3\left(x-1\right)+8}{x-1}\)

\(=\dfrac{3\left(x-1\right)}{x-1}+\dfrac{8}{x-1}=3+\dfrac{8}{x-1}\)

Biểu thức nguyên khi \(\dfrac{8}{x-1}\) nguyên 

⇒ 8 ⋮ x - 1

⇒ x - 1 ∈ Ư(8) 

⇒ x - 1 ∈ {1; -1; 2; -2; 4; -4; 8; -8}

⇒ x ∈ {2; 0; 3; -1; 5; -3; 9; -7} 

\(\left(2x+1\right)^2=\dfrac{25}{4}\)

\(\Rightarrow\left(2x+1\right)^2=\left(\pm\dfrac{5}{2}\right)^2\)

+) \(2x+1=\dfrac{5}{2}\)

\(\Rightarrow2x=\dfrac{3}{2}\)

\(\Rightarrow x=\dfrac{3}{4}\)

+) \(2x+1=-\dfrac{5}{2}\)

\(\Rightarrow2x=-\dfrac{7}{2}\)

\(\Rightarrow x=-\dfrac{7}{4}\)

(2x + 1)² = 25/4

2x + 1 = 5/2 hoặc 2x + 1 = -5/2

*) 2x + 1 = 5/2

2x = 5/2 - 1

2x = 3/2

x = 3/2 : 2

x = 3/4

*) 2x + 1 = -5/2

2x = -5/2 - 1

2x = -7/2

x = -7/2 : 2

x = -7/4

Vậy x = -7/4; x = 3/4

1/2 . 2ⁿ⁺⁴ . 2 = 2⁵⁴

2ⁿ⁺⁴⁺ⁿ = 2⁵⁴ . 2

2²ⁿ⁺⁴ = 2⁵⁵

2n + 4 = 55

2n = 55 - 4

2n = 51

n = 51/2

a) Diện tích xung quanh bể:

(50 + 90) . 2 . 80 = 22400 (cm²)

Thể tích bể:

50 . 90 . 80 = 360000 (cm³)

b) 0,2 m = 20 cm

Thể tích nước đã bơm vào bể:

50 . 90 . (80 - 20) = 270000 (cm³) = 27 (m³)

diện tích xung quanh  của bể  là

\([\)(90+50) x2\(]\) x 80 = 22400 cm² 

thể tích bể là

90 x 50 x 80 =360000 cm³ 

đổi 0,2 m = 20 cm

chiều co của bể khi bơm nước xong là

80 - 20 = 60 cm

thể tích  nước đã bơm vào bể  là

90 x 50 x 60 =270000 cm³

(\(\dfrac{12}{25}\))^{\(^x\)}  = (\(\dfrac{5}{3}\))^-2 - (-\(\dfrac{3}{5}\))⁴

(\(\dfrac{12}{25}\))^\(x\)  = (\(\dfrac{3}{5}\))² - (\(\dfrac{3}{5}\))⁴

(\(\dfrac{12}{25}\))^{\(^x\)} = \(\dfrac{9}{25}\) - \(\dfrac{81}{625}\)

(\(\dfrac{12}{25}\))^{\(^x\)}  = \(\dfrac{144}{625}\)

(\(\dfrac{12}{25}\))^{\(^x\)} = (\(\dfrac{12}{25}\))²

       \(x\) = 2

(12/25)ˣ = (5/3)⁻² - (-3/5)⁴

(12/25)ˣ = 9/25 - 81/625

(12/25)ˣ = 144/625

(12/25)ˣ = (12/25)²

x = 2

a) 1/20 - (x - 8/5) = 1/10

x - 8/5 = 1/20 - 1/10

x - 8/5 = -1/20

x = -1/20 + 8/5

x = 31/20

b) 7/4 - (x + 5/3) = -12/5

x + 5/3 = 7/4 + 12/5

x + 5/3 = 83/20

x = 83/20 - 5/3

x = 149/60

c) x - [17/2 - (-3/7 + 5/3)] = -1/3

x - (17/2 - 26/21) = -1/3

x - 305/42 = -1/3

x = -1/3 + 305/42

x = 97/14

a) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\left(-\dfrac{7}{12}\right)\cdot1\dfrac{2}{5}\)

\(\Rightarrow\dfrac{1}{6}x=\left(-\dfrac{7}{12}\right)\cdot\dfrac{7}{5}\)

\(\Rightarrow\dfrac{1}{6}x=-\dfrac{49}{60}\)

\(\Rightarrow x=-\dfrac{49}{60}:\dfrac{1}{6}\)

\(\Rightarrow x=-\dfrac{49}{10}\) 

b) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)

\(\Rightarrow\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\left(\pm\dfrac{3}{2}\right)^2\)

+) \(\dfrac{1}{5}-\dfrac{3}{2}x=\dfrac{3}{2}\)

\(\Rightarrow\dfrac{3}{2}x=\dfrac{1}{5}-\dfrac{3}{2}\)

\(\Rightarrow\dfrac{3}{2}x=-\dfrac{13}{10}\)

\(\Rightarrow x=-\dfrac{13}{10}:\dfrac{3}{2}\)

\(\Rightarrow x=-\dfrac{13}{15}\)

+) \(\left(1,25-\dfrac{4}{5}x\right)^3=-125\)

\(\Rightarrow\left(\dfrac{5}{4}-\dfrac{4}{5}x\right)^3=\left(-5\right)^3\)

\(\Rightarrow\dfrac{5}{4}-\dfrac{4}{5}x=-5\)

\(\Rightarrow\dfrac{4}{5}x=\dfrac{5}{4}+5\)

\(\Rightarrow\dfrac{4}{5}x=\dfrac{25}{4}\)

\(\Rightarrow x=\dfrac{25}{4}:\dfrac{4}{5}\)

\(\Rightarrow x=\dfrac{125}{16}\)

a, \(\dfrac{2}{3}\)\(x\) - \(\dfrac{1}{2}\)\(x\) = (- \(\dfrac{7}{12}\)). 1\(\dfrac{2}{5}\)

    \(x\).(\(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)) = (- \(\dfrac{7}{12}\)) . \(\dfrac{7}{5}\)

    \(x\). \(\dfrac{1}{6}\) = - \(\dfrac{49}{60}\)

    \(x\)      = - \(\dfrac{49}{60}\).6

    \(x\)      = -\(\dfrac{49}{10}\)

16 + 4x = 2 ÷ 2

=> 16 + 4x = 1

=> 4x = 1 - 16

=> 4x = -15

=> x = -15/4

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

\(2A=2\cdot\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\)

\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{101}}\)

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

\(A=1-\dfrac{1}{2^{100}}\)

b) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2023\cdot2024}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2023}-\dfrac{1}{2024}\)

\(=1-\dfrac{1}{2024}\)

\(=\dfrac{2024}{2024}-\dfrac{1}{2024}\)

\(=\dfrac{2023}{2024}\)

cứu