\(\Leftrightarrow\left(x^2-4x+6\right)\cdot\left(x^2-4x+10\right)=21\)
\(\Leftrightarrow\left(x^2-4x+6\right)\cdot\left(x^2-4x+10\right)-21=0\)
\(\Leftrightarrow x^4-4x^3+10x^2-4x^3+16x^2-40x+6x^2-24x+60-21=0\)
\(\Leftrightarrow x^4-8x^3+32x^2-64x+39=0\)
\(\Leftrightarrow x^4-x^3-7x^3+7x^2+25x^2-25x-39x+39=0\)
\(\Leftrightarrow x^3\left(x-1\right)-7x^2\cdot\left(x-1\right)+25x\left(x-1\right)-39x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\cdot\left(x^3-7x^2+25x-39\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3-3x^2-4x^2+12x+13x-39\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x-3\right)-4x\cdot\left(x-3\right)+13\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x^2-4x+13\right)=0\)
\(\hept{\begin{cases}x-1=0\\x-3=0\\x^2-4x+13=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\x=3\\x\notin R\end{cases}}\)
Vậy phương trình của tập nghiệm là S={1;3}
