$c)(2x+3)^2-3(x-4)(x+4)=(x-2)^2$
$\to [(2x+3)^2-(x-2)^2]-3(x^2-16)=0$
$\to (2x+3-x+2)(2x+3+x-2)-3x^2+48=0$
$\to (x+5)(3x+1)-3x^2+48=0$
$\to 3x^2+16x+5-3x^2+48=0$
$\to 16x+53=0$
$\to x=\dfrac{-53}{16}$
Vậy S={$\dfrac{-53}{16}$}
$d)x(x+1)-(x-3)(x-4)=5x$
$\to x^2+x-x^2+7x-12-5x=0$
$\to 3x-12=0$
$\to x=4$
Vậy $S={4}$
$e) \dfrac{x}{10}-(\dfrac{x}{30}+\dfrac{2x}{45})=\dfrac{4}{5}$
$\to 9x-3x-4x-72=0$
$\to 2x-72=0$
$\to x=36$
Vậy $S={36}$
