Đáp án:
$a) (3x+4)^2 -(3x+1)(3x+1)=49$
$⇔9x^2+24x+16 - 9x^2 -6x-1=49$
$⇔18x +15 =49$
$⇔18x = 49 -15$
$⇔18x=34$
$⇔x = 34 : 18$
$⇔x=\dfrac{17}{9}$
Vậy $x=\dfrac{17}{9}$
$b) (x-1)^3 +3(x+1)^2 =(x^2-2x+4)(x+2)$
$ ⇔ x^3 -3x^2+3x-1 +3(x^2+2x+1) =x^3+2x^2-2x^2+-4x+4x+8$
$ ⇔x^3 -3x^2+3x -1 +3x^2+6x+3 =x^3+8$
$⇔x^3 -x^3 -3x^2+3x^2+3x+6x=8-3+1$
$⇔9x = 4$
$⇔x=4 : 9$
$⇔x=\dfrac{2}{3}$
Vậy $x=\dfrac{2}{3}$