Đặt \(x^2-x+2=a,\, x^2=b\)
\(\to a^2-3ab+2b^2=0\\\leftrightarrow a^2-2ab-ab+2b^2=0\\\leftrightarrow a(a-2b)-b(a-2b)=0\\\leftrightarrow (a-b)(a-2b)=0\\\leftrightarrow a-b=0\,\, or\,\, a-2b=0\\\leftrightarrow x^2-x+2-x^2=0\,\, or\,\, x^2-x+2-2x^2=0\\\leftrightarrow 2-x=0\,\, or\,\, -x^2-x+2=0\\\leftrightarrow x=2\,\, or -x^2-2x+x+2=0\\\leftrightarrow x=2\,\, or\,\, -x(x+2)+(x+2)=0\\\leftrightarrow x=2\,\, or\,\, (1-x)(x+2)=0\\\leftrightarrow x=2\,\, or\,\, 1-x=0\,\, or\,\, x+2=0\\\leftrightarrow x=2\,\, or\,\, x=1\,\, or\,\, x=-2\)
Vậy pt có tập nghiệm \(S=\{-2;1;2\}\)
