\(P=5x^2-2x^3+6x^4+2x^3-2x^4-3x^4+2\)
\(P=5x^2+x^4+2\)
Đặt t = x2 (t >= 0)
\(P=t^2+5t+2\)
\(\left[{}\begin{matrix}t=\dfrac{-5+\sqrt{17}}{2}\left(loai\right)\\\dfrac{-5-\sqrt{17}}{2}\left(loai\right)\end{matrix}\right.\)
Vậy P(x) = 0 vô nghiệm
a: Ta có: \(P\left(x\right)=5x^2-2x^3+6x^4+2x^3-2x^4-3x^4+2\)
\(=x^4+5x^2+2\)
\(P\left(1\right)=1^4+5\cdot1^2+2=8\)
\(P\left(-1\right)=\left(-1\right)^4+5\cdot\left(-1\right)^2+2=8\)
\(P\left(0\right)=0^4+5\cdot0^2+2=2\)
\(P\left(\dfrac{1}{5}\right)=\dfrac{1}{625}+5\cdot\dfrac{1}{25}+2=\dfrac{1376}{625}\)
b: Ta có: \(x^4\ge0\forall x\)
\(5x^2\ge0\forall x\)
Do đó: \(x^4+5x^2\ge0\forall x\)
\(\Leftrightarrow x^4+5x^2+2>0\forall x\)
Vậy: Đa thức P(x) không có nghiệm
