Giải thích các bước giải:

a.$(x+1)(x^2+5x-2)-(x^3+1)=0$

$\to x^3+6x^2+3x-2-x^3-1=0$

$\to 6x^2+3x-3=0$

$\to 2x^2+x-1=0$

$\to (2x-1)(x+1)=0$

$\to x\in\{\dfrac12, -1\}$

b.$x^2+(x+2)(11x+7)=14$

$\to 12x^2+29x+14=14$

$\to 12x^2+29x=0$

$\to x(12x+29)=0$
$\to x\in\{0, -\dfrac{29}{12}\}$

c.$x^3+1=x(x+1)$

$\to (x+1)(x^2-x+1)=x(x+1)$

$\to (x+1)(x^2-x+1)-x(x+1)=0$

$\to (x+1)(x^2-x+1-x)=0$

$\to (x+1)(x^2-2x+1)=0$

$\to (x+1)(x-1)^2=0$

$\to x\in\{-1, 1\}$

d.$x^3+x^2+x+1=0$

$\to x^2(x+1)+(x+1)=0$

$\to (x+1)(x^2+1)=0$

Vì $x^2+1>0$

$\to x+1=0$

$\to x=-1$

e.$x^2-3x+2=0$

$\to (x-1)(x-2)=0$

$\to x\in\{1, 2\}$

f.$-x^2+5x-6=0$

$\to x^2-5x+6=0$

$\to (x-2)(x-3)=0$

$\to x\in\{2, 3\}$

g.$4x^2-12x+5=0$

$\to (2x-5)(2x-1)=0$

$\to x\in\{\dfrac52, \dfrac12\}$

h.$2x^2+5x+3=0$

$\to (2x+3)(x+1)=0$

$\to x\in\{-\dfrac32, -1\}$