(100.x-10).(6+x)=0
⇒ \(\left[ \begin{array}{l}100x-10=0 ⇔ 100x =10 ⇔ x = \frac{10}{100} = 0,1 \\6+x=0 ⇔ x=-6\end{array} \right.\)
(6.x+12).(15-5.x)=0
⇒ \(\left[ \begin{array}{l}6×x +12=0 ⇔ 6×x =-12 ⇔ x = -2 \\15-5×x=0 ⇔ 5×x = 15 ⇔ x = 3 \end{array} \right.\)
x.(x+7)-15.(x+7)=0
= (x-15) . (x+7) = 0
⇒ \(\left[ \begin{array}{l}x-15=0 ⇔ x =15 \\x+7=0 ⇔ x = -7 \end{array} \right.\)