Nhận thấy \(x=0\) không phải nghiệm
\(\Leftrightarrow\left(x+1\right)\left(x+8\right)\left(x+2\right)\left(x+4\right)=28x^2\)
\(\Leftrightarrow\left(x^2+8+9x\right)\left(x^2+8+6x\right)=28x^2\)
\(\Leftrightarrow\left(\frac{x^2+8+9x}{x}\right)\left(\frac{x^2+8+6x}{x}\right)=28\)
\(\Leftrightarrow\left(x+\frac{8}{x}+9\right)\left(x+\frac{8}{x}+6\right)-28=0\)
Đặt \(x+\frac{8}{x}+6=a\) ta được:
\(\left(a+3\right).a-28=0\)
\(\Rightarrow a^2+3a-28=0\Rightarrow\left[{}\begin{matrix}a=4\\a=-7\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{8}{x}+6=4\\x+\frac{8}{x}+6=-7\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+2x+8=0\\x^2+13x+8=0\end{matrix}\right.\) \(\Leftrightarrow...\)
