lam lại, bài trên có lỗi kĩ thuật .-.
ĐKXĐ: x - 2 khác 0 => x khác 2
\(\left(x^4-2x^2-8\right):\left(x-2\right)=0\)
\(\Leftrightarrow x^4-2x^2-8=0\)
\(\Leftrightarrow\left(x^2\right)^2-2x^2\cdot1+1-9=0\)
\(\Leftrightarrow\left(x^2-1\right)^2-9=0\)
\(\Leftrightarrow\left(x^2-1\right)^2=9\Leftrightarrow\left[{}\begin{matrix}x^2-1=3\\x^2-1=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=4\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\x^2=-2\left(loai\right)\end{matrix}\right.\)
Vậy pt có 2 nghiệm là \(\left\{{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
ĐKXĐ: x - 2 \(\ne0\Rightarrow x\ne2\)
(\(x^4-2x^2-8\))\(:\left(x-2\right)=0\)
\(\Rightarrow x^4-2x^2-8=0\)
\(\Rightarrow\left(x^2\right)^2-2x^2\cdot1+1-9=0\)
\(\Rightarrow\left(x^2-1\right)^2=7\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-1=7\\x^2-1=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x^2=8\Rightarrow x=\pm\sqrt{8}\\x^2=-6\left(loai\right)\end{matrix}\right.\)
Vậy pt có 2 no là \(\left\{{}\begin{matrix}x=\sqrt{8}\\x=-\sqrt{8}\end{matrix}\right.\)
