(y² - 2y + 1) - 9 = 0
⇔ (y - 1)² - 3² = 0
⇔ (y - 1 - 3)(y - 1 + 3) = 0
⇔ (y - 4)(y + 2) = 0
⇔ y - 4 = 0 hoặc y + 2 = 0
*) y - 4 = 0
⇔ y = 4
*) y + 2 = 0
⇔ y = -2
Vậy S = {-2; 4}
\(\left(y^2-2y+1\right)-9=0\\ \Leftrightarrow\left(y-1\right)^2-3^2=0\\ \Leftrightarrow\left(y-1-3\right).\left(y-1+3\right)=0\\ \Leftrightarrow\left(y-4\right).\left(y+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}y-4=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=4\\y=-2\end{matrix}\right.\\ Vậy:S=\left\{-2;4\right\}\)
\(\left(y^2-2y+1\right)-9=0\\ \Leftrightarrow y^2-2y-8=0\\ \Leftrightarrow y^2-4y+2y-8=0\\ \Leftrightarrow\left(y^2-4y\right)+\left(2y-8\right)=0\\ \Leftrightarrow y\left(y-4\right)+2\left(y-4\right)=0\\ \Leftrightarrow\left(y+2\right)\left(y-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}y+2=0\\y-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}y=-2\\y=4\end{matrix}\right.\)
Vậy phương trình có nghiệm \(S=\left\{-2;4\right\}.\)
