\(f\left(x\right)=\left(\frac{1}{2x^2}+1\right)^2+4x^2\)
Ta có
\(\left(a+b\right)^2=a^2+2ab+b^2\)
\(\Rightarrow a^2+b^2=\left(a+b\right)^2-2ab\)
\(\Rightarrow f\left(x\right)=\left(\frac{1}{2x^2}+1+4x^2\right)^2-\left(\frac{1}{2x^2}+1\right)4x^2\)
Đa thức có ngiệm khi
\(\left(\frac{1}{2x^2}+1+4x^2\right)^2-\left(\frac{1}{2x^2}+1\right)4x^2=0\)
\(\Rightarrow\left(\frac{1}{2x^2}+1+4x^2\right)^2-\left(\frac{4x^2}{2x^2}+4x^2\right)=0\)
\(\Rightarrow\left(\frac{1}{2x^2}+1+4x^2\right)^2-\left(2+4x^2\right)=0\)
\(\Rightarrow\frac{1}{4x^4}+1+16x^4+\frac{1}{2x^2}+4x^2+2-2-4x^2=0\)
\(\Rightarrow\frac{1}{4x^4}+1+16x^4+\frac{1}{2x^2}=0\)
Mà \(\frac{1}{4x^2}\ge0\) ; \(1>0\) ; \(16x^4\ge0\) ; \(\frac{1}{2x^2}\ge0\)
=> đa thức vô nghiệm
