Verified answer

Ответ:

[tex]1) ~~ x_1 = 0 ~~ ; ~~ x_2 = -3\dfrac{2}{3}[/tex]

[tex]2) ~~ x_{1} = \cfrac{\sqrt{35}}{2} } ~~ ; ~~ x_2 =- \cfrac{\sqrt{35}}{2} }[/tex]

Объяснение:

[tex]\hspace{-1,5em}1) ~ (3x-1)(x+4) = -4 \\\\ 3x^2 - x+ 12x \bold{- 4}= \bold{-4} \\\\ 3x^2+11x = 0 \\\\ x(3x+11) =0[/tex]

Приравняем каждый множитель  к нулю

[tex]1) ~x_1 = 0[/tex]

[tex]\hspace{-1,1em}2) ~ 3x+ 11 =0 \\\\ 3x = - 11 \\\\ x_2=-\cfrac{11}{3} = -3\cfrac{2}{3}[/tex]


[tex]\hspace{-1,4em}2)~ (2x-1)^2 -6(6-x) = 2x \\\\ 4x^2-4x+1 -36+6x =2x \\\\ 4x^2-4x-2x+6x -35 =0 \\\\ 4x^2-6x+6x -35 =0 \\\\ 4x^2 -35 =0 \\\\ 4x^2 = 35 \\\\ x^2 = \cfrac{35}{4} \\\\ x_{1;2} = \pm \sqrt{\dfrac{35}{4} } \\\\\ x_{1;2} = \pm \cfrac{\sqrt{35}}{2} }[/tex]