Bài 1: 

a: BCNN(10;12)=60

b: BCNN(24;10)=120

c: BCNN(4;14;26)=364

d: BCNN(6;8;10)=120

Ta có:

\(x\) ⋮ 10 và \(x\) ⋮ 6 

\(\Rightarrow x\in B\left(10\right)\) và \(x\in B\left(6\right)\)

\(\Rightarrow x\in BC\left(6;10\right)\)

Mà: \(B\left(10\right)=\left\{0;10;20;30;40;50;60;70;80;90;100;110;120;...\right\}\)

\(B\left(6\right)=\left\{0;6;12;18;24;30;36;42;48;54;60;66;72;...\right\}\)

\(\Rightarrow BC\left(6;10\right)=\left\{0;30;60;90;120;150;180;...\right\}\)

Lại có: \(100< x< 150\) 

\(\Rightarrow x\in\left\{120\right\}\) 

`@` `\text {Ans}`

`\downarrow`

`a)`

`x + 10 = 20`

`=> x = 20 -10`

`=> x = 10`

Vậy, `x = 10`

`b)`

`2 * x + 15 = 35`

`=> 2x = 35 - 15`

`=> 2x = 20`

`=> x = 20 \div 2`

`=> x = 10`

Vậy, `x = 10`

`c)`

`3 * ( x + 2 ) = 15`

`=> x + 2 = 15 \div 3`

`=> x + 2 = 5`

`=> x = 5 - 2`

`=> x = 3`

Vậy, `x = 3`

`d)`

`10 * x + 15 * 11 = 20 * 10`

`=> 10x + 165 = 200`

`=> 10x = 200 - 165`

`=> 10x = 35`

`=> x = 35 \div 10`

`=> x = 3,5`

Vậy,` x = 3,5`

`e)`

`4 * ( x + 2 ) = 3 * 4`

`=> x + 2 = 12 \div 4`

`=> x + 2 = 3`

`=> x = 3 - 2`

`=> x = 1`

Vậy,` x = 1`

`f)`

`33 x + 135 = 26 * 9`

`=> 33x + 135 = 234`

`=> 33x = 234 - 135`

`=> 33x = 99`

`=> x = 99 \div 33`

`=> x = 3`

Vậy, `x = 3`

`g)`

`2 * x + 15 + 16 + 17 = 100`

`=> 2x + 48 = 100`

`=> 2x = 100 - 48`

`=> 2x = 52`

`=> x = 52 \div 2`

`=> x =26`

`h)`

`2 * (x + 9 + 10 + 11) = 4 . 12 . 25`

`=> 2 * (x + 9 + 10 + 11) = 4*25*12`

`=> 2 * (x + 9 + 10 + 11) = 100*12`

`=> x + 9 + 10 + 11 = 100*12 \div 2`

`=> x + 30 = 600`

`=> x = 600 - 30`

`=> x = 570`

Vậy, `x = 570.`

a) \(x+10=20\Leftrightarrow x=10\)

b) \(2x+15=35\Leftrightarrow2x=20\Leftrightarrow x=10\)

c) \(3.\left(x+2\right)=15\Leftrightarrow x+2=5\Leftrightarrow x=3\)

d) \(10x+15.11=20.10\Leftrightarrow10x+165=200\Leftrightarrow10x=35\Leftrightarrow x=\dfrac{35}{10}=\dfrac{7}{2}\)

e) \(4.\left(x+2\right)=3.4\Leftrightarrow x+2=3\Leftrightarrow x=1\)

f) \(35x+135=26.9\Leftrightarrow35x=234-135\Leftrightarrow35x=99\Leftrightarrow x=\dfrac{99}{35}\)

g) \(2x+15+16+17=100\Leftrightarrow2x+48=100\Leftrightarrow2x=52\Leftrightarrow x=26\)

h) \(2.\left(x+9+10+11\right)=4.12.25\)

\(\Leftrightarrow x+30=2.12.25\)

\(\Leftrightarrow x=600-30\)

\(\Leftrightarrow x=570\)

Số đó là 120 em nhé!

x⋮10; x⋮12; x⋮15

⇒x∈BC(10;12;15)

10=5.2

12=2².3

15=3.5

BCNN(10;12;15)=5.2².3=60

BC(10;12;15)={0;60;120;...}

mà 100<x<150 nên x=120

Số cần tìm là 120

Từ gt ta có: x \(\in BC\left(10,12,15\right)\left\{60;120;180;...\right\}\)  Mà  100<x<150

Vậy x =120

Bài 2:

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

=>\(x\in\left\{2;0;3;-1;4;-2;7;-5\right\}\)

b: \(\Leftrightarrow x+3\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)

=>\(x\in\left\{-2;-4;0;-6;2;-8;12;-18\right\}\)

c: \(\Leftrightarrow x+3\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)

=>\(x\in\left\{-2;-4;-1;-5;0;-6;1;-7;3;-9;9;-15\right\}\)

d: =>x+1+15 chia hết cho x+1

=>\(x+1\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)

=>\(x\in\left\{0;-2;2;-4;4;-6;14;-16\right\}\)

x chia het cho 10,12,15

suy ra xla bc (10;12;15)

bcnn (60)= (0;60;120;180;240)

do 100<x<150 suy ra xthuoc (120)

vay x =120

x chia hết cho 10, 12, 15

=> x thuộc BC(10, 12, 15)

Ta có: 10 = 2.5; 12 = 2².3; 15 = 3.5

=> BCNN(10, 12, 15) = 2².3.5 = 60

=> x thuộc BC(10, 12, 15) = B(60) = {0; 60; 120; 180; ...}

Mà 100 < x < 150

Vậy x = 120.

x=120 vì 120/10=12,120/12=10,120/15=8

ĐS 120

Ta có : 10=2.5 ; 12=2².3 ; 15 = 3.5

BCNN(10,12,15) = 2².3.5 = 60

=> BC(10,12,15) = B(60) = {0,60,120,240,...}

Vì 100<x<150 nên x=120

Giải :

Vì : x chia hết cho 10 

      x chia hết cho 12 

     x chia hết cho 15

=>  x là BCNN ( 10 , 12 , 15)

10 = 2 * 5 

\(12=2^2 ×3\)

15= 3 * 5 

BCNN ( 10 , 12 , 15) = 2^2 * 3 * 5 = 60 

BC  (10 , 12 , 15 ) = B( 60 ) = { 0 , 60 , 120 , 180 , 240 , 300 , .....}

Mà  100 < x < 150 

=> x thuộc 160 

Vậy số cần tìm là 160.

^^