2:

a: =>x^2+3x-4x-12-(x^2-5x+x-5)=8

=>x^2-x-12-x^2+4x+5=8

=>3x-7=8

=>3x=15

=>x=5

b: =>3x^2+3x-2x-2-3x^2-21x=13

=>-20x=15

=>x=-3/4

c: =>x^2-25-x^2-2x=9

=>-2x=25+9=34

=>x=-17

d: =>x^3-1-x^3+3x=1

=>3x-1=1

=>3x=2

=>x=2/3

Lời giải:
|5(3x+1)|+|2(3x+1)|+|3x+1|=4$

$5|3x+1|+2|3x+1|+|3x+1|=4$

$(5+2+1)|3x+1|=4$

$8|3x+1|=4$

$|3x+1|=\frac{1}{2}$

$3x+1=\pm \frac{1}{2}$

$\Rightarrow x=\frac{-1}{6}$ hoặc $x=\frac{-1}{2}$

b, \(\left(x-5\right)\left(x-4\right)-\left(x+1\right)\left(x-2\right)=7\)

\(\Rightarrow x^2-9x+20-x^2+x+2=7\)

\(\Rightarrow-8x+22=7\)

\(\Rightarrow-8x=-15\)

\(\Rightarrow x=\frac{15}{8}\)

c, \(\left(3x-4\right)\left(x-2\right)=3x\left(x-9\right)-3\)

\(\Rightarrow3x^2-10x+8=3x^2-27x-3\)

\(\Rightarrow3x^2-10x-3x^2+27x=\left(-3\right)+\left(-8\right)\)

\(\Rightarrow17x=-11\)

\(\Rightarrow x=-\frac{11}{17}\)

d, \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(5-x^2\right)=6x\)

\(\Rightarrow x^3+3x^2+9x-3x^2-9x-27+5x-x^3=6x\)

\(\Rightarrow6x=-27\)

\(\Rightarrow x=-\frac{27}{6}\)

\(\Rightarrow x=-\frac{9}{2}\)

e, \(\left(3x-5\right)\left(x+1\right)-\left(3x-1\right)\left(x+1\right)=x-4\)

\(\Rightarrow3x^2-2x-5-3x^2-2x+1=x-4\)

\(\Rightarrow-4=x-4\)

\(\Rightarrow x=0\)

b)    (x - 5)(x - 4) - (x + 1)(x - 2) = 7
<=> x² - 9x + 20 - x² + x + 2 - 7 = 0
<=> 8x - 15 = 0 <=> x = 15/8

c)    (3x - 4)(x - 2) = 3x(x - 9) - 3
<=> 3x² - 10x + 8 = 3x² - 27x - 3
<=> 17x = -11 <=> x = -11/17

d)    (x - 3)(x² + 3x + 9) + x(5 - x²) = 6x
<=> x³ - 27 - x³ + 5x - 6x = 0
<=> x = -27

e)    (3x - 5)(x + 1) - (3x - 1)(x + 1) = x - 4
<=> (x + 1)(3x - 5 - 3x + 1) - x + 4 = 0
<=> -4x - 4 - x + 4 = 0 <=> x = 0

a) \(\left(2x-3\right)\left(2x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=3\\2x=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=1\\x=2\end{matrix}\right.\)

c) \(2x\left(3x-1\right)-3x\left(5+2x\right)=0\)

\(\Rightarrow x\left[2\left(3x-1\right)-3\left(5+2x\right)\right]=0\)

\(\Rightarrow x\left(6x-2-15-6x\right)\)

\(\Rightarrow-16x=0\)

\(\Rightarrow x=0\)

d) \(\left(3x-2\right)\left(3x+2\right)-4\left(x-1\right)=0\)

\(\Rightarrow9x^2-4-4x+4=0\)

\(\Rightarrow9x^2-4x=0\)

\(\Rightarrow x\left(9x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\9x-4=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\end{matrix}\right.\)

\(a,\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ b,\left(x-4\right)\left(x-1\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\\x=2\end{matrix}\right.\)

\(c,2x\left(3x-1\right)-3x\left(5+2x\right)=0\\ \Leftrightarrow6x^2-2x-15x-6x^2=0\\ \Leftrightarrow-17x=0\\ \Leftrightarrow x=0\\ d,\left(3x-2\right)\left(3x+2\right)-4\left(x-1\right)=0\\ \Leftrightarrow9x^2-4-4x+4=0\\ \Leftrightarrow9x^2-4x=0\\ \Leftrightarrow x\left(9x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\9x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\end{matrix}\right.\)

a) 3/2.|x - 5/3| - 4/5 = 4/3.|x - 5/3| + 1

<=> 3/2.|x - 5/3| = 4/3.|x - 5/3| + 1 + 4/5

<=> 3/2.|x - 5/3| = 9/5 + 4|x - 5/3|/3

<=> 3/2.|x - 5/3| - 4.|x - 5/3|/3 = 9/5

<=> |x - 5/3|/6 = 9/5

<=> |x - 5/3| = 9/5.6

<=> |x - 5/3| = 54/5

<=> x - 5/3 = 54/5 hoặc x - 5/3 = -54/5

       x = 54/5 + 5/3         x = -54/5 - 5/3

       x = 187/15              x = -137/15

b) 2.|3x + 1| = 1/3.|3x + 1| + 5

<=> 2.|3x + 1| - 1/3.|3x + 1| = 5

<=> 5/3.|3x + 1| = 5

<=> 5.|3x + 1| = 5.3

<=> 5.|3x + 1| = 15

<=> |3x + 1| = 15 : 5

<=> |3x + 1| = 3

       3x + 1 = 3 hoặc 3x + 1 = -3 

       3x = 3 - 1           3x = -3 - 1

       3x = 2                3x = -4

       x = 2/3               x = -4/3

=> x = 2/3 hoặc x = -4/3

c) làm tương tự câu a) mình hơi lời

Làm câu c) cho

\(\frac{1}{4}-\frac{5}{2}\left|3x-\frac{1}{5}\right|=\frac{2}{3}\left|3x-\frac{1}{5}\right|-\frac{2}{3}\)

\(\Leftrightarrow\frac{1}{4}+\frac{2}{3}=\frac{2}{3}\left|3x-\frac{1}{5}\right|+\frac{5}{2}\left|3x-\frac{1}{5}\right|\)

\(\Leftrightarrow\frac{3}{12}+\frac{8}{12}=\left|3x-\frac{1}{5}\right|\left(\frac{2}{3}+\frac{5}{2}\right)\)

\(\Leftrightarrow\left|3x-\frac{1}{5}\right|\left(\frac{4}{6}+\frac{15}{6}\right)=\frac{11}{12}\)

\(\Leftrightarrow\frac{19}{6}\left|3x-\frac{1}{5}\right|=\frac{11}{12}\)

\(\Leftrightarrow\left|3x-\frac{1}{5}\right|=\frac{11}{12}.\frac{6}{19}\)

\(\Leftrightarrow\left|3x-\frac{1}{5}\right|=\frac{11}{38}\)

\(\Leftrightarrow\orbr{\begin{cases}3x-\frac{1}{5}=\frac{11}{38}\\3x-\frac{1}{5}=-\frac{11}{38}\end{cases}}\)

Giải tiếp nha

\(\Leftrightarrow4x-3x^2+20-15x-9x^2-12x-4+\left(3x+2\right)^3=8x^3-1\)

\(\Leftrightarrow-12x^2-23x+16+27x^3+54x^2+36x+8=8x^3-1\)

\(\Leftrightarrow27x^3+42x^2+13x+24-8x^3+1=0\)

\(\Leftrightarrow19x^3+42x^2+12x+25=0\)