a/ Nhận thấy \(x=0\) không phải nghiệm, chia cả 2 vế của pt cho \(x^2\):
\(x^2+5x-10+\frac{10}{x}+\frac{4}{x^2}=0\)
\(\Leftrightarrow x^2+\frac{4}{x^2}+5\left(x+\frac{2}{x}\right)-10=0\)
Đặt \(x+\frac{2}{x}=a\Rightarrow x^2+4+\frac{4}{x^2}=a^2\Rightarrow x^2+\frac{4}{x^2}=a^2-4\)
Phương trình trở thành:
\(a^2-4+5a-10=0\)
\(\Leftrightarrow a^2+5a-14=0\) \(\Rightarrow\left[{}\begin{matrix}a=2\\a=-7\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{2}{x}=2\\x+\frac{2}{x}=-7\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-2x+2=0\left(vn\right)\\x^2+7x+2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{-7+\sqrt{41}}{2}\\x=\frac{-7-\sqrt{41}}{2}\end{matrix}\right.\)
b/ \(x^4-8x^2+x+12=0\)
\(\Leftrightarrow x^4-8x^2+16+x-4=0\)
\(\Leftrightarrow\left(x^2-4\right)^2+x-4=0\)
Đặt \(x^2-4=a\Rightarrow-4=a-x^2\)
Phương trình trở thành:
\(a^2+x+a-x^2=0\)
\(\Leftrightarrow\left(a-x\right)\left(a+x\right)+x+a=0\)
\(\Leftrightarrow\left(a-x+1\right)\left(x+a\right)=0\)
\(\Leftrightarrow\left(x^2-4-x+1\right)\left(x+x^2-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-3=0\\x^2+x-4=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{1\pm\sqrt{13}}{2}\\x=\frac{-1\pm\sqrt{17}}{2}\end{matrix}\right.\)
