a,
$A=B$
$↔(x-3)(x+4)-2(3x-2)=(x-4)^2$
$↔x^2+x-12-6x+4=x^2-8x+16$
$↔x^2-5x-8=x^2-8x+16$
$↔x^2-5x-8-x^2+8x-16=0$
$↔(x^2-x^2)+(-5x+8x)+(-8-16)=0$
$↔3x-24=0$
$↔3x=24$
$↔x=8$
Vậy $x=8$ thì $A=B$
$----------$
b,
$A=B$
$↔(x+2)(x-2)+3x^2=(2x+1)^2+2x$
$↔x^2-4+3x^2=4x^2+4x+1+2x$
$↔4x^2-4=4x^2+6x+1$
$↔4x^2-4-4x^2-6x-1=0$
$↔(4x^2-4x^2)+(-4-1)-6x=0$
$↔-6x-5=0$
$↔-6x=5$
$↔x=-\dfrac56$
Vậy $x=-\dfrac56$ thì $A=B$
$----------$
c,
$A=B$
$↔(x-1)(x^2+x+1)-2x=x(x-1)(x+1)$
$↔x^3-1-2x=x(x^2-1)$
$↔x^3-2x-1=x^3-x$
$↔x^3-2x-1-x^3+x=0$
$↔(x^3-x^3)+(-2x+x)-1=0$
$↔-x-1=0$
$↔-x=1$
$↔x=-1$
Vậy $x=-1$ thì $A=B$
$----------$
d,
$A=B$
$↔(x+1)^3-(x-2)^3=(3x-1)(3x+1)$
$↔[(x+1)-(x-2)][(x+1)^2+(x+1)(x-2)+(x-2)^2]=9x^2-1$
$↔(x+1-x+2)(x^2+2x+1+x^2-x-2+x^2-4x+4)=9x^2-1\\\leftrightarrow3(3x^2-3x+3)=9x^2-1\\\leftrightarrow9x^2-9x+9=9x^2-1\\\leftrightarrow9x^2-9x+9-9x^2+1=0\\\leftrightarrow(9x^2-9x^2)-9x+(9+1)=0\\\leftrightarrow-9x+10=0\\\leftrightarrow-9x=-10\\\leftrightarrow x=\dfrac{10}{9}$
Vậy $x=\dfrac{10}{9}$ thì $A=B$