Ответ:

[tex]x_1 = -3\dfrac{2}{7} ~ ~; ~~ x_2 = -1[/tex]

Объяснение:

[tex]|1-3x| + 4|x+3|= 12[/tex]

[tex]1 -3x = 0 ~~~~ ; ~~~~ 3+x =0 \\\\ x =\dfrac{1}{3} ~~~~~~~~~~;~~~~ x = -3[/tex]

[tex]\setlength{\unitlength}{23mm}\begin{picture}(1,1) \linethickness{0.2mm} \put(0.9,-0.2) {\sf -3} \put(0 ,0.09){ \Large ~~~~~I } \put(1.02 ,0.09){ \Large ~~~~~II } \put(2 ,0.09){ \Large ~~~~~III } \put(1,0) {\line(0,2){0.3}} \put(1,0.3) {\line(1,0){2}} \put(2,-0.26) {\sf $\dfrac{1}{3} $} \put(2.05,0) {\line(0,2){0.3}} \put(1,0.3) {\line(-1,0){1} } \ \put(0,0){\vector (1,0){3}} \put(2.94,-0.15){\sf x} \end{picture}[/tex]

    +                            +                           —                  1 - 3x

    —                           +                            +                  x + 3

[tex]\hspace{-1,4em}\text{I} )~~ |1-3x| +4\cdot |x+3| = 12 \\\\ 1 -3x -4 (x+3) = 12 \\\\ -7x = 23 \\\\ x _1= -3\dfrac{2}{7} ~ \checkmark ~ ~ , ~ x \in (- \infty ~ ; ~ -3 )[/tex]

[tex]\hspace{-1,6em}\text{II} )~~ |1-3x| +4\cdot |x+3| = 12 \\\\ 1 -3x +4 (x+3) = 12 \\\\ x+13 = 12 \\\\ x_2 = -1 ~ \checkmark ~ ~ , ~ x \in [-3 ~ ; ~ 1/3 )[/tex]

[tex]\hspace{-1,9em}\text{III} )~~ |1-3x| +4\cdot |x+3| = 12 \\\\ -(1 -3x) +4 (x+3) = 12 \\\\ 7x+11 = 12 \\\\ x_2 = \dfrac{1}{7} ~ \varnothing ~ ~ , ~ x \in [ 1/3 ~; ~ \infty )[/tex]