\(x^2-2x+1=25\)
\(\Leftrightarrow\)\(\left(x-1\right)^2=25\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-1=5\\x-1=-5\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)
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b) \(3\left(x-1\right)^2-3x\left(x-5\right)=1\)
\(\Leftrightarrow\)\(3x^2-6x+3-3x^2+15x=1\)
\(\Leftrightarrow\)\(9x=-2\)
\(\Leftrightarrow\)\(x=-\frac{2}{9}\)
Vậy...