\(P=x^4-2x^2-3\left|x^2-1\right|-9\)
\(P=x^4-2x^2+1-3\left|x^2-1\right|-10\)
\(P=\left(x^2-1\right)^2-3\left|x^2-1\right|-10\)
\(P=\left|x^2-1\right|^2-3\left|x^2-1\right|-10\)
Đặt: \(\left|x^2-1\right|=t\ge0\) ta có:
\(pt\Leftrightarrow t^2-3t-10=t^2-3t+\dfrac{9}{4}-\dfrac{49}{4}\)
\(=\left(t-\dfrac{3}{2}\right)^2-\dfrac{49}{4}\ge\left(-\dfrac{3}{2}\right)^2-\dfrac{49}{4}=\dfrac{9}{4}-\dfrac{49}{4}=-10\)
Dấu "=" khi: \(t=0\Leftrightarrow\left|x^2-1\right|=0\Leftrightarrow x=\pm1\)
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