Bài 1: 

c: x=4

b: x=2

a) (x - 140) : 7 = 3³ - 2³ . 3

(x - 140) : 7 = 27 - 8 . 3 = 27 - 24 = 3

x - 140 = 3 x 7 = 21

x = 21 + 140 = 161

b) x³ . x² = 2⁸ : 2³

x⁵ = 2⁵

=> x = 2

c) (x + 2) . ( x - 4) = 0

x = -2 hoặc 4

d) 3^x-3 - 3² = 2 . 3² =

3^x-3 - 9 = 2 . 9 = 18

3^x-3 = 18 + 9 = 27

3^x-3 = 3³

=> x - 3 = 3

x = 3 + 3 = 6

2.

a) 9 : ( x + 2 )

9 ⋮ 1 ; 9 ⋮ 3 ; 9 ⋮ 9

=> x = -1 ; 1 ; 7

b: 30 chia hết cho x

45 chia hết cho x

Do đó: \(x\inƯC\left(30;45\right)=Ư\left(15\right)\)

mà x>10

nen x=15

c: \(\Leftrightarrow x+2\in\left\{1;-1;3;-3;9;-9\right\}\)

hay \(x\in\left\{-1;-3;1;-5;7;-11\right\}\)

d: =>x+3+14 chia hết cho x+3

=>\(x+3\in\left\{1;-1;2;-2;7;-7;14;-14\right\}\)

hay \(x\in\left\{-2;-4;-1;-5;4;-10;11;-17\right\}\)

a) (x - 1)² = 1.

<=> x - 1 = 1 hoặc x - 1 = -1.

<=> x = 2 hoặc x = 0.      

b) 7^{2x - 6} = 49.

<=> 7^{2x - 6} = 7².   

<=> 2x - 6 = 2.

<=> x = 4.  

c) (2x - 16)⁷ = 128.

<=> (2x - 16)⁷ = 2⁷.

<=> 2x - 16 = 2.

<=> x = 9.

a: \(\Leftrightarrow x-3=-13\)

hay x=-10

a: \(x\in\left\{-6;-5;-4;-3;-2;-1;0;1;2;3\right\}\)

c: \(x\in\left\{-4;-3;-2;-1\right\}\)

a) 8 chia hết cho x + 1

--> x + 1 là ước của 8.

TH1: x + 1 = 8 

TH2: x + 1 = 4

TH3: x + 1 = 2

TH4: x + 1 = 1

Giải ra được x = 7; x = 3; x = 1; x = 0

\(a,\Rightarrow12x-91=101\\ \Rightarrow12x=192\\ \Rightarrow x=16\\ b,\Rightarrow x:23+45=133\\ \Rightarrow x:23=88\\ \Rightarrow x=\dfrac{88}{23}\\ c,\Rightarrow\left(6x-39\right):7=3\\ \Rightarrow6x-39=21\\ \Rightarrow6x=60\\ \Rightarrow x=10\\ d,\Rightarrow3x-24=\dfrac{148}{73}\\ \Rightarrow3x=\dfrac{1900}{73}\\ \Rightarrow x=\dfrac{1900}{219}\\ e,\Rightarrow\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\\ f,\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\\ d,\left(9-x\right)^3=64=4^3\\ \Rightarrow9-x=4\\ \Rightarrow x=5\\ h,\Rightarrow x=27\\ i,\Rightarrow6x=312\cdot12=624\cdot6\\ \Rightarrow x=624\\ j,\Rightarrow\left(19x+104\right):14=25-42=-17\\ \Rightarrow19x+104=-238\\ \Rightarrow19x=-342\\ \Rightarrow x=-18\)

Bài 4:

\(a,\Rightarrow\left(x+2\right)\left(y+1\right)=3\cdot7=7\cdot3=21\cdot1=1\cdot21\)

Vậy \(\left(x;y\right)\in\left\{\left(19;0\right);\left(1;6\right);\left(5;2\right)\right\}\)