Bài 2: 

a: Ta có: \(2^{x+1}\cdot3^y=12^x\)

\(\Leftrightarrow2^{x+1}\cdot3^y=2^{2x}\cdot3^x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1=2x\\x=y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

a: \(x+7⋮x+2\)

=>\(x+2+5⋮x+2\)

=>\(5⋮x+2\)

=>\(x+2\in\left\{1;-1;5;-5\right\}\)

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

b: \(2x+5⋮x+1\)

=>\(2x+2+3⋮x+1\)

=>\(3⋮x+1\)

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

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

c: \(3x-2⋮x+3\)

=>\(3x+9-11⋮x+3\)

=>\(-11⋮x+3\)

=>\(x+3\in\left\{1;-1;11;-11\right\}\)

=>\(x\in\left\{-2;-4;8;-14\right\}\)

d: \(12x+1⋮3x+2\)

=>\(12x+8-7⋮3x+2\)

=>\(-7⋮3x+2\)

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

=>\(3x\in\left\{-1;-3;5;-9\right\}\)

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

e: \(x^2+3x+5⋮x+3\)

=>\(x\left(x+3\right)+5⋮x+3\)

=>\(5⋮x+3\)

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

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

f: \(x^2-2x+3⋮x+2\)

=>\(x^2+2x-4x-8+11⋮x+2\)

=>\(11⋮x+2\)

=>\(x+2\in\left\{1;-1;11;-11\right\}\)

=>\(x\in\left\{-1;-3;9;-13\right\}\)

l) (x + 9) . (x² – 25) = 0  

<=> (x + 9) . (x – 5) . (x + 5) = 0   

<=> \(\left[{}\begin{matrix}\text{x + 9 = 0}\\x-5=0\\x+5=0\end{matrix}\right.\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)

Vậy S = \(\left\{-9,5,-5\right\}\)

      e) |x - 4 |< 7         

<=> \(\left[{}\begin{matrix}x-4=7\\x-4=-7\end{matrix}\right.< =>\left[{}\begin{matrix}x=11\\x=-3\end{matrix}\right.\)

Vậy S = \(\left\{11;-3\right\}\)

I,(x+9).(x^2-25)=0

tương đương:x+9=0

                       x^2-25=0

tương đương : x=-9

                       x=5

e,\(\left|x-4\right|\)=7

tương đương x-4=4

                       x-4=-4

tương đương :x=0

                        x=-8

a, 75 - 5 . ( 15 - 10 ) - ( -60 ) = 75 - 5 . 5 + 60 = 135 - 25 = 110

b, 136 - ( -7 ) + 6 - 23 - 36 = 136 + 7 + 6 - 23 - 36 = ( 136 - 36 ) + ( 7 + 6 - 23 ) = 100 - 10 = 90

c, ( -15 - 17 ) . ( -15 + 7 ) = -32 . ( -8 ) = 256

d, ( -9 - 18 ) : ( -9 ) = -27 : ( -9 ) = 3

Bài 2:

a, 2x - 17 = ( -3x - 18 )

<=> 2x + 3x = 17 - 18

<=> 5x = -1

<=> x = -1/5

b, ( 13 - IxI ) + 15 = 20

<=> 13 - |x| = 5

<=> |x| = 8

<=> x = 8 hoặc x = -8  

c, Ix + 5I - ( x + 5 ) = 0

<=> |x + 5| = x + 5 

Đk: x + 5 ≥ 0 => x ≥ -5

\(\Leftrightarrow\orbr{\begin{cases}x+5=x+5\\x+5=-x-5\end{cases}\Leftrightarrow}\orbr{\begin{cases}0x=0\\2x=-10\end{cases}\Leftrightarrow}\orbr{\begin{cases}x\ge-5\\x=-5\end{cases}}\)

Vậy  x ≥ -5

b: UCLN(18;42)=6

1/ 3(x - 2) = 9 => x - 2 = 3 => x = 5

2/ 3(x - 36) = 216 => x - 36 = 72 => x = 108

3/ (x - 3)³ = 3³ => x - 3 = 3 => x = 6

4/ 2^x = 32 => 2^x = 2⁵ => x = 5

5/ 3x - 16 = 2.7⁴ : 7³ = 14 => 3x = 30 => x = 10

6/ (6x - 39) : 3 = 201 => 6x - 39 = 603 => 6x = 642 => x = 107

7/ (3x - 15)² = 81 => (3x - 15)² = 9² => 3x - 15 = 9 => 3x = 24 => x = 8

8/ 3(x + 4) = 105 => x + 4 = 35 => x = 31

9/ 12x - 4³ = 4.8⁴ : 8³ = 32 => 12x = 32 : 64 => 12x = 1/2 => x = 1/24

10/ 41 - (2x - 5) = 680 => 2x - 5 = -639 => 2x = -634 => x = - 317