(x - 1)2 + (x - 2)2 = 1 (1)
\(\Leftrightarrow\) x2 - 2x + 1 + x2 - 4x + 4 - 1 = 0
\(\Leftrightarrow\) 2x2 - 6x + 4 = 0
\(\Leftrightarrow\) 2(x2 - 3x + 2) = 0
\(\Leftrightarrow\) x2 - 3x + 2 = 0
\(\Leftrightarrow\) x2 - 2x - x + 2 = 0
\(\Leftrightarrow\) x(x - 2) - (x - 2) = 0
\(\Leftrightarrow\) (x - 2)(x - 1) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
Vậy S = {2; 1}
x4 - 3x3 + 3x2 - 3x + 2 = 0 (2)
\(\Leftrightarrow\) x4 - 3x3 + 3x2 - x - 2x + 2 = 0
\(\Leftrightarrow\) x(x3 - 3x2 + 3x - 1) - 2(x - 2) = 0
\(\Leftrightarrow\) x(x - 1)3 - 2(x - 1) = 0
\(\Leftrightarrow\) (x - 1)[x(x - 1) - 2] = 0
\(\Leftrightarrow\) (x - 1)(x2 - x - 2) = 0
\(\Leftrightarrow\) (x - 1)(x2 - 2x + x - 2) = 0
\(\Leftrightarrow\) (x - 1)[x(x - 2) + (x - 2)] = 0
\(\Leftrightarrow\) (x - 1)(x - 2)(x + 1) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-1\end{matrix}\right.\)
Vậy S = {1; 2; -1}
x3 - 7x + 6 = 0 (3)
\(\Leftrightarrow\) x3 - x - 6x + 6 = 0
\(\Leftrightarrow\) x(x2 - 1) - 6(x - 1) = 0
\(\Leftrightarrow\) x(x - 1)(x + 1) - 6(x - 1) = 0
\(\Leftrightarrow\) (x - 1)[x(x + 1) - 6] = 0
\(\Leftrightarrow\) (x - 1)(x2 + x - 6) = 0
\(\Leftrightarrow\) (x - 1)(x2 + x + \(\frac{1}{4}\) - \(\frac{25}{4}\)) = 0
\(\Leftrightarrow\) (x - 1)[(x + \(\frac{1}{2}\))2 - \(\frac{25}{4}\)] = 0
\(\Leftrightarrow\) (x - 1)(x + \(\frac{1}{2}\) - \(\frac{5}{2}\))(x + \(\frac{1}{2}\) + \(\frac{5}{2}\)) = 0
\(\Leftrightarrow\) (x - 1)(x - 2)(x + 3) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-3\end{matrix}\right.\)
Vậy S = {1; 2; -3}
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