a, (x - 1)(2x² - 10) = 0
⇔\(\left[ \begin{array}{l}x - 1=0\\2x² - 10=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x =1\\x² =5\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x =1\\x=+- √5\end{array} \right.\)
KL:..
b, (2x - 7)² - 6(2x - 7)(x - 3) = 0
⇔(2x - 7).[2x - 7-6(x-3)]=0
⇔\(\left[ \begin{array}{l}2x - 7=0\\2x - 7-6x+18=0end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=7/2\\11=4x\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x =7/2\\x=11/4\end{array} \right.\)