54\(⋮\) \(x\);   72 ⋮ \(x\);     90 ⋮ \(x\) ⇒ \(x\) \(\in\) ƯC(54; 72; 90) ⇒ \(x\)\(\in\)ƯCLN(54;72;90)

54 = 2.3²;        72 = 2³.3²; 90 = 2.3².5

ƯCLN(54; 72; 90) = 2.3² = 18

\(x\in\)Ư(18) = { 1; 2; 3; 6; 9; 18}

Ta có:

\(100=2^2.5^2\)\(;150=2.3.5^2\)

\(ƯCLN\left(100;150\right)=2.5^2=50\)

Ta có:

\(35=5.7;75=3.5^2\)

\(ƯCLN\left(35;75\right)=5\)

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

A là số nguyên => x - 1 \(\in\) Ư(8) = {-8,-4,-2,-1,1,2,4,8}

Ta có bảng:

Vậy với giá trị x = 9 sẽ là giá trị x lớn nhất để A là số nguyên.

`@` `\text {Ans}`

`\downarrow`

\(x\div\dfrac{1}{5}+x\div\dfrac{1}{7}+x\div\dfrac{1}{9}=42\)

\(\Rightarrow x\cdot5+x\cdot7+x\cdot9=42\)

\(\Rightarrow x\cdot\left(5+7+9\right)=42\)

\(\Rightarrow x\cdot21=42\)

\(\Rightarrow x=42\div21\)

\(\Rightarrow x=2\)

Vậy, `x = 2.`

a, 117600 = 2⁵.3.5².7² = (2⁴.3) . (2.5²).7² = 48.50.49 = 48.49.50

b, 132600 = 2³.3.5².13.17 = (2.25).(3.17).(2².13) = 50.51.52

a) \(117600=48.49.50\)

b) \(132600=50.51.52\)

\(B=\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{26\cdot29}\)

\(B=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{29}\)

\(B=\dfrac{1}{2}-\dfrac{1}{29}\)

\(B=\dfrac{27}{58}\)

B= 3/2x5 + 3/5x8+ 3/8x11 + ... + 3/26x29

B= 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/26 - 1/29

B= 1/2-1/29

B=27/58

90 - [ 120 - ( 13 - 8)³]

 = 90 - [ 120 -  5³ ]

= 90 - [ 120 - 125]

= 90 - (-5)

= 95

\(A=\dfrac{3n+4}{n-1}\inℤ\)

\(\Rightarrow3n+4⋮n-1\)

\(\Rightarrow3n+4-3\left(n-1\right)⋮n-1\)

\(\Rightarrow3n+4-3n+3⋮n-1\)

\(\Rightarrow7⋮n-1\)

\(\Rightarrow n-1\in\left\{-1;1;-7;7\right\}\)

\(\Rightarrow n\in\left\{0;2;-6;8\right\}\)

Cô tick xanh cho em nhưng Nguyễn Đức TRí lần sau em làm nhớ thêm đkxđ vào nhá