1.

a,  \(x-14=3x+18\)                                                                       

\(\Rightarrow x-3x=18+14\)                                                                 

\(\Rightarrow-2x=32\Rightarrow x=\frac{32}{-2}=-16\)

b, \(\left(x+7\right).\left(x-9\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+7=0\\x-9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-7\\x=9\end{cases}}}\)

c, \(\left|2x-5\right|-7=22\)                                                                     

\(\Rightarrow\left|2x-5\right|=22+7\)

\(\Rightarrow\left|2x-5\right|=29\)

\(\Rightarrow\orbr{\begin{cases}2x+5=29\\2x-5=29\end{cases}}\Rightarrow\orbr{\begin{cases}2x=24\\2x=34\end{cases}\Rightarrow}\orbr{\begin{cases}x=12\\x=17\end{cases}}\)

d,  \(\left(\left|2x\right|-5\right)-7=22\)

\(\Rightarrow\left(\left|2x\right|-5\right)=29\)

\(\Rightarrow\left|2x\right|=29+5\Rightarrow\left|2x\right|=34\Rightarrow x=\pm17\)

e, \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\)

Vì \(\left|x+3\right|\ge0;\left|x+9\right|\ge0;\left|x+5\right|\ge0;4x\ge0\)

Nên \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\ge0\)

\(\Rightarrow\left|x+3\right|>0\Rightarrow\left|x+3\right|=x+3\)

     \(\left|x+9\right|>0\Rightarrow\left|x+9\right|=x+9\)

      \(\left|x+5\right|>0\Rightarrow\left|x+5\right|=x+5\)

Ta có : 

\(x+3+x+9+x+5=4x\)

\(\Rightarrow3x+\left(3+9+5\right)=4x\)

\(\Rightarrow4x-3x=17\)

\(\Rightarrow x=17\)

2. a , b sai đề bn 

c, \(\left(5x+1\right).\left(y-1\right)=4\)

\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)

\(\text{ }Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)

Ta có bảng sau : 

d, \(5xy-5x+y=5\)

\(\Rightarrow\left(5xy-5x\right)+y=5\)

\(\Rightarrow5x.\left(y-1\right)+y=5\)

\(\Rightarrow\left(5x+1\right).\left(y-1\right)=4\)

\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)

\(Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)

Ta có bảng sau : 

x - 14 = 3x + 18

x - 3x = 18 + 14

-2x= 32

x= 32 : (-2)

x=-16

a, \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}\Rightarrow}\orbr{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)

b. \(\left(x^2+1\right)\left(x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^2+1=0\\x-4=0\end{cases}\Rightarrow}\orbr{\begin{cases}x^2=-1\left(Voly\right)\\x=4\end{cases}\Rightarrow x=4}\)

c, \(2x^2-\frac{1}{3}x=0\)

\(\Leftrightarrow x\left(2x-\frac{1}{3}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\2x-\frac{1}{3}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{6}\end{cases}}\)

d, \(\left(\frac{4}{5}\right)^{5x}=\left(\frac{4}{5}\right)^7\)

\(\Rightarrow5x=7\)

\(\Rightarrow x=\frac{7}{5}\)

e, Ta có: \(A=\frac{x+5}{x-2}=\frac{\left(x-2\right)+7}{x-2}=1+\frac{7}{x-2}\)

Để A ∈ Z <=> (x - 2) ∈ Ư(7) = { ±1; ±7 }

 Vậy....

a) \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)

Vậy : ....

b) \(\left(x^2+1\right)\left(x-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x-4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=-1\left(loại\right)\\x=4\end{cases}}\)

c) \(2x^2-\frac{1}{3}x=0\)

\(\Leftrightarrow x\left(2x-\frac{1}{3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x-\frac{1}{3}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{6}\end{cases}}\)

Vậy :...

a) 2x - 5 = -19

2x= -19+5

2x= -14

x= -14:2

x= -7

Vậy...

b) |x + 2| = 17

TH1: x+2= -17

x= -17 -2

x= -19

TH2: x+2=17

x= 17-2

x= 15

Vậy...

c) c)(x + 5).(x - 9) = 0

=> ( x+5) = 0 hoặc ( x-9) =0

Nếu x+5= 0 thì x= 0-5 => x = -5

Nếu x-9 = 0 thì x = 0+9=> x =9

Vậy...

d) 5x -14 = x-34

5x -14 -x = 34

4x-14= -34

4x= -34+14

4x= - 20

x= -20:4

x= -5

Vậy...

B1:

a) \(\left(10x+9\right)x-\left(5x-1\right)\left(2x+3\right)=8\)

\(10x^2+9x-10x^2-15x+2x+3-8=0\)

\(-4x-5=0\)

\(-4x=5\Leftrightarrow x=-\dfrac{5}{4}\)

b) \(\left(3x-5\right)\left(7-5x\right)+\left(5x+2\right)\left(3x-2\right)-2=0\)

\(21x-15x^2-35+25x+15x^2-10x+6x-4-2=0\)

\(42x-41=0\)

\(x=\dfrac{41}{42}\)

3.

\(x=\left|2\right|\Rightarrow x=\pm2\)

Thay x = 2 vào A ta có:

A = (3.2+5)(2.2+1) + (4.2+1)(5.2+2)

= 11.5 + 9.12

= 55 + 108

= 163

Thay x = -2 vào A ta có:

A = (-2.3+5)(-2.2+1) + (-2.4+1)(-2.5+2)

= (-1)(-3) + (-7)(-8)

= 3 + 56

= 59

Thay x = -1 vào B ta có:

B = (-1-3)(-1+7) - (-1.2-5)(-1-1)

= (-4).6 - (-7)(-2)

= -24 - 14

= -38

Vậy \(A=163\Leftrightarrow x=2\)

\(A=59\Leftrightarrow x=-2\)

\(B=-38\Leftrightarrow x=-1\)

a)  ( 3x - 1 ) ( 2x + 7 )  - ( x + 1 ) ( 6x + 5 ) = 16 

<=> 6x² + 21x - 2x - 7 - ( 6x² - 5x + 6x - 5) = 16

<=> 6x² + 21x - 2x - 7 - ( 6x² + x - 5 )        = 16 

<=> 6x²+ 21x - 2x - 7 - 6x² -x + 5              = 16 

<=> 18x - 2                                             = 16 

<=>  18x                                                 = 18 

=>        x                                                 = 1

Vậy....  

\(a,-12\left(x-5\right)+7\left(3-x\right)=5\)

\(-12x+60+21-7x=5\)

\(-12x-7x=5-60-21\)

\(-19x=-76\Leftrightarrow x=4\)

\(b,30\left(x+2\right)-6\left(x-5\right)-24x=100\)

\(30x+60-6x+30-24x=100\)

\(30x-6x-24x=100-60-30\)

\(0x=10\left(vl\right)\)

Vậy pt vô nghiệm 

\(c,\left(x-23\right):14+25=42-1^{2002}\)

\(\left(x-23\right).\frac{1}{14}+25=42-1\)

\(\frac{x}{14}-\frac{23}{14}+25=41\)

\(\frac{x}{14}-\frac{23}{14}=16\)

\(\frac{x}{14}=\frac{247}{14}\)

\(x=247\)