Bài 1: 

a: UCLN(12;52)=4

UC(12;52)={1;2;4}

a, bổ sung đề 

 \(\dfrac{29-x}{21}+1+\dfrac{27-x}{23}+1+\dfrac{25-x}{25}+1+\dfrac{23-x}{27}+1+\dfrac{21-x}{29}+1=0\)

\(\Leftrightarrow\dfrac{50-x}{21}+\dfrac{50-x}{23}+\dfrac{50-x}{25}+\dfrac{50-x}{27}+\dfrac{50-x}{29}=0\)

\(\Leftrightarrow\left(50-x\right)\left(\dfrac{1}{21}+\dfrac{1}{23}+\dfrac{1}{25}+\dfrac{1}{27}+\dfrac{1}{29}\ne0\right)=0\Leftrightarrow x=50\)

Giải:

Bài 1:

a) \(6^2\): 4 . 3 + 2 . \(5^2\) 

= 36 : 4 . 3 + 2 . 25

= 27 + 50

= 77

77 = 7 . 11

b) 5 . \(4^2\) - 18 : \(3^2\)

= 5 . 16 - 18 : 9

= 80 - 2

= 78

78 = 2 .3 . 13

Bài 2: 

a) Ta có: \(70⋮x\)

\(84⋮x\)

Do đó: \(x\inƯC\left(70;84\right)\)

\(\Leftrightarrow x\inƯ\left(14\right)\)

\(\Leftrightarrow x\in\left\{1;2;7;14\right\}\)

mà x>8

nên x=14

x ⋮ 12 , x ⋮ 25 , x ⋮ 30 và 0 < x < 500

Vì x ⋮ 12 , x ⋮ 25 và x ⋮ 30 nên x ∈ BC(12; 25; 30)

Ta có: 12 = 22.3; 25 = 52 và 30 = 2.3.5

BCNN(12; 25; 30) = 22. 3. 52 = 300

BC(12; 25; 30) = {0; 300; 600; ...}

Vì 0 < x < 500 nên x = 300.

a: Ta có: \(30⋮x\)

\(\Leftrightarrow x\in\left\{1;-1;2;-2;3;-3;5;-5;6;-6;10;-10;15;-15;30;-30\right\}\)

mà x<8

nên \(x\in\left\{1;2;3;5;6;-1;-2;-3;-5;-6;-10;-15;-30\right\}\)

b: Ta có: \(70⋮x\)

\(84⋮x\)

Do đó: \(x\inƯC\left(70;84\right)\)

\(\Leftrightarrow x\in\left\{1;2;7;14;-1;-2;-7;-14\right\}\)

mà x>8

nên x=14

c: Ta có: \(90⋮x\)

\(126⋮x\)

Do đó: \(x\inƯC\left(90;126\right)\)

\(\Leftrightarrow x\in\left\{1;-1;2;-2;3;-3;6;-6;9;-9;18;-18\right\}\)

mà x>9

nên x=18

d: Ta có: \(x⋮12\)

\(x⋮25\)

\(x⋮30\)

Do đó: \(x\in BC\left(12;25;30\right)\)

\(x\in\left\{...;-300;0;300;600;...\right\}\)

mà 0<x<500

nên x=300