a/ \(x^4-4x^2+4+x^2-6x+9=0\)
\(\Leftrightarrow\left(x^2-2\right)^2+\left(x-3\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=2\\x=3\end{matrix}\right.\) \(\Rightarrow\) pt vô nghiệm
b/ Nhận thấy \(x=0\) ko phải nghiệm, chia 2 vế cho \(x^2\)
\(3x^2+\frac{3}{x^2}-5x-\frac{5}{x}+8=0\)
\(\Leftrightarrow3\left(x+\frac{1}{x}\right)^2-5\left(x+\frac{1}{x}\right)+2=0\)
\(\Leftrightarrow\left(x+\frac{1}{x}-1\right)\left(3x+\frac{3}{x}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{1}{x}-1=0\\3x+\frac{3}{x}-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=0\\3x^2-2x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\frac{1}{x}\right)^2+\frac{3}{4}=0\\3\left(x-\frac{1}{x}\right)^2+\frac{8}{3}=0\end{matrix}\right.\)
Cả 2 pt đều vô nghiệm nên pt đã cho vô nghiệm
