$1) (x+1)^2+3(x+1)-4=0$
$\to (x+1)^2+3(x+1)+\dfrac{9}{4}-\dfrac{9}{4}-4=0$
$\to [(x+1)^2+3(x+1)+(\dfrac{3}{2}^2)-\dfrac{25}{4}=0$
$\to (x+1+\dfrac{3}{2})^2-\dfrac{5}{2}^2=0$
$\to (x + 1 +\dfrac{3}{2}-\dfrac{5}{2})(x+1+\dfrac{3}{2}+\dfrac{5}{2})=0$
$\to x(x+5)=0$
$\to x=0$
$\to x+5=0 \to x=-5$
Vậy $S={0,-5}$
2) $(x^2-2x)^2-(x^2-2x)-6=0$
$\to (x^2-2x)^2-(x^2-2x)+ \dfrac{1}{2}^2-\dfrac{1}{2}^2-6=0$
$\to (x^2-2x)^2-(x^2-2x)+\dfrac{1}{4}-\dfrac{25}{4}=0$
$\to (x^2-2x-\dfrac{1}{2})^2-\dfrac{5}{2}^2=0$
$\to (x^2-2x-\dfrac{1}{2}-\dfrac{5}{2})(x^2-2x-\dfrac{1}{2}+\dfrac{5}{2})=0$
$\to (x^2-2x -3)(x^2-2x+2)=0$
Do: $x^2-2x+2=0\to$ vô nghiệm
$\to x^2 -2x-3=0$
$\to x^2 -3x +x -3=0$
$\to x(x-3)+(x-3)=0$
$\to (x-3)(x+1)=0$
$\to x=3; x=-1$
Vậy $S={3;-1}$
3)$(x^2+2x+3)(x^2+2x+1)=3$
$\to (x^2+2x+3)[(x^2+2x+3)-2]-3=0$
$\to (x^2+2x+3)^2-2(x^2+2x+3)-3=0$
$\to [(x^2+2x+3)^2-2(x^2+2x+3)+1]-4=0$
$\to (x^2+2x+3-1)^2-2^2=0$
$\to(x^2+2x+2-2)(x^2+2x+2+2)=0$
$\to(x^2+2x)(x^2+2x+4)=0$
Do $x^2+2x=0 \to x= 0; -2$
$\to x^2+2x+4=0$
$\to$ vô nghiệm
Vậy $S={0;-2}$
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