Đáp án:
1/ $(x^2-x+2)^2-(2x+2)^2=0\\↔(x^2-x+2-2x-2)(x^2-x+2+2x+2)=0\\<=>(x^2-3x)(x^2+x+4)=0\\<=>x(x-3)(x^2+x+4)=0$
$x^2+x+4\\=x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{15}{4}\\=\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}$
Nhận thấy: $\left(x+\dfrac{1}{2}\right)^2\ge 0$
$<=>\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}\ge \dfrac{15}{4}>0$
$<=>x^2+x+4>0$
mà $x(x-3)(x^2+x+4)=0$
$→\left[\begin{array}{1}x=0\\x-3=0\end{array}\right.\\↔\left[\begin{array}{1}x=0\\x=3\end{array}\right.$
Vậy $S=\{0;3\}$
2/ Đặt $x^2+2x-1=y$
$=>y(y-1)=2$
$↔y^2-y-2=0\\<=>y^2-2y+y-2=0\\↔(y^2-2y)+(y-2)=0\\↔y(y-2)+(y-2)=0\\<=>(y+1)(y-2)=0\\<=>(x^2+2x-1+1)(x^2+2x-1-2)=0\\↔(x^2+2x)(x^2+3x-x-3)=0\\↔x(x+2)[(x^2+3x)-(x+3)]=0\\↔x(x+2)[x(x+3)-(x+3)]=0\\↔x(x+2)(x-1)(x+3)=0\\↔\left[\begin{array}{1}x=0\\x+2=0\\x-1=0\\x+3=0\end{array}\right.\\↔\left[\begin{array}{1}x=0\\x=-2\\x=1\\x=-3\end{array}\right.$
Vậy $S=\{0;-2;1;-3\}$
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