Ta có: \(x^2-6x-2+\frac{14}{x^2-6x+7}=0\)
\(\Leftrightarrow\frac{\left(x^2-6x-2\right)\left(x^2-6x+7\right)+14}{x^2-6x+7}=0\)
\(\Leftrightarrow x^4-12x^3+41x^2-30x-14+14=0\)
\(\Leftrightarrow x^4-12x^3+41x^2-30x=0\)
ĐKXĐ : \(x^2-6x+7\ne0\)
=> \(x^2-6x+9-2\ne0\)
=> \(\left(x-3\right)^2\ne2\)
=> \(\left[{}\begin{matrix}x-3\ne-\sqrt{2}\\x-3\ne\sqrt{2}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x\ne3-\sqrt{2}\\x\ne3+\sqrt{2}\end{matrix}\right.\)
- Ta có : \(x^2-6x-2+\frac{14}{x^2-6x+7}=0\)
Đặt : \(a=x^2-6x+7\)
=> \(a-9=x^2-6x-2\)
- Thay \(a-9=x^2-6x-2\), \(a=x^2-6x+7\) vào phương trình ta được : \(a-9+\frac{14}{a}=0\)
=> \(\frac{a^2}{a}-\frac{9a}{a}+\frac{14}{a}=0\)
=> \(a^2-9a+14=0\)
=> \(a^2-7a-2a+14=0\)
=> \(a\left(a-2\right)-7\left(a-2\right)=0\)
=> \(\left[{}\begin{matrix}a-7=0\\a-2=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}a=7\\a=2\end{matrix}\right.\)
- Thay \(a=x^2-6x+7\) vào phương trình trên ta được :
\(\left[{}\begin{matrix}x^2-6x+7=7\\x^2-6x+7=2\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x^2-6x=0\\x^2-6x=5\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x\left(x-6\right)=0\\x^2-5x-x-5=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x\left(x-6\right)=0\\x\left(x-1\right)-5\left(x-1\right)=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x\left(x-6\right)=0\\\left(x-1\right)\left(x-5\right)=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=0\\x-6=0\\x-5=0\\x-1=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=0\\x=6\\x=5\\x=1\end{matrix}\right.\) ( TM )
Vậy phương trình có nghiệm là x = 0, x = 6, x = 5, x = 1 .
