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)  ( 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....  

`Answer:`

\(3\left(\frac{5}{3}x-7\right)-2\left(1.5x+6\right)-\left(5-x\right)\left(x+4\right)=80+x^2\)

\(\Leftrightarrow3\left(\frac{5x}{3}-7\right)-2\left(5x+6\right)-\left(5-x\right)\left(x+4\right)=80+x^2\)

\(\Leftrightarrow5x-21-10x-12-5x-20+x^2+4x=80+x^2\)

\(\Leftrightarrow5x-21-10x-12-5x-20+4x=80\)

\(\Leftrightarrow-6x-53=80\)

\(\Leftrightarrow-6x=133\)

\(\Leftrightarrow x=-\frac{133}{6}\)

\(\frac{4}{5}x^2\left(\frac{x}{3}-\frac{1}{2}\right)-\left(\frac{1}{5}x-\frac{2}{3}\right)\left(4\frac{x^2}{3}+1\right)=\frac{22}{45}x^2\)

\(\Leftrightarrow36x^2\left(\frac{x}{3}-\frac{1}{2}\right)-45\left(\frac{x}{5}-\frac{2}{3}\right)\left(\frac{4x^2}{3}+1\right)=22x^2\)

\(\Leftrightarrow12x^3-18x^2-12x^3-9x+40x^2+30=22x^2\)

\(\Leftrightarrow22x^2-9x+30=22x^2\)

\(\Leftrightarrow-9x+30=0\)

\(\Leftrightarrow-9x=-30\)

\(\Leftrightarrow x=\frac{10}{3}\)

a, \(3.\left(\dfrac{5}{3}x-7\right)-2\left(1,5x+6\right)-\left(5-x\right)\left(x+4\right)=80+x^2\)

\(\Rightarrow5x-21-3x-12-\left(5x+20-x^2-4x\right)-x^2=80\)

\(\Rightarrow5x-21-3x-12-5x-20+x^2+4x-x^2=80\)

\(\Rightarrow5x-3x-5x+4x+x^2-x^2=80+21+12+20\)

\(\Rightarrow x=133\)

Câu b tương tự! Cứ tách ra!

a) \(3\left(\dfrac{5}{3}x-7\right)-2\left(1,5x+6\right)-\left(5-x\right)\left(x+4\right)=80+x^2\) (1)

\(\Leftrightarrow\left(5x-21\right)-\left(3x+12\right)-\left(5x+20-x^2-4x\right)=80+x^2\)

\(\Leftrightarrow5x-21-3x-12-5x-20+x^2+4x=80+x^2\)

\(\Leftrightarrow x-53+x^2=80+x^2\)

\(\Leftrightarrow x+x^2-x^2=80+53\)

\(\Leftrightarrow x=133\)

Vậy tập nghiệm phương trình (1) là \(S=\left\{133\right\}\)

b) chưa rõ đề.