\(\left(x-3\right)^4-3\left(x^2-6x+10\right)=1\)
\(\Leftrightarrow x^4-12x^3+51x^2-90x+51=1\)
\(\Leftrightarrow x^4-12x^3+51x^2-90x+51-1=0\)
\(\Leftrightarrow x^4-12x^3+51x^2-90x+50=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-5\right)\left(x^2-6x+10\right)=0\)
vì \(x^2-6x+10\ne0\) nên:
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
Vậy: Phương trình có tập nghiệm là: S = {1; 5}
Ta có: \(\left(x-3\right)^4-3\left(x^2-6x+10\right)=1\)
\(\Leftrightarrow\left(x^2-6x+9\right)^2-3x^2+18x-30-1=0\)
\(\Leftrightarrow x^4+36x^2+81+12x^3+18x^2+108x-3x^2+18x-31=0\)
\(\Leftrightarrow x^4+12x^3+51x^2+126x+50=0\)
