Nhận thấy \(x=0\) không phải nghiệm
\(\left(x+3\right)\left(x+4\right)\left(x+2\right)\left(x+6\right)=14x^2\)
\(\Leftrightarrow\left(x^2+12+7x\right)\left(x^2+12+8x\right)=14x^2\)
\(\Leftrightarrow\left(x+\frac{12}{x}+7\right)\left(x+\frac{12}{x}+8\right)=14\)
Đặt \(x+\frac{12}{x}+7=a\)
\(a\left(a+1\right)-14=0\)
\(\Leftrightarrow a^2+a-14=0\Rightarrow\left[{}\begin{matrix}a=\frac{-1-\sqrt{57}}{2}\\a=\frac{-1+\sqrt{57}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{12}{x}+7=\frac{-1-\sqrt{57}}{2}\\x+\frac{12}{x}+7=\frac{-1+\sqrt{57}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2+\frac{15-\sqrt{57}}{2}x+12=0\\x^2+\frac{15+\sqrt{57}}{2}x+12=0\end{matrix}\right.\)
Số xấu quá, bạn tự giải nốt
