\(\left(x-4\right)^2-9=0\)
\(\Leftrightarrow\left(x-4\right)^2-3^2=0\)
\(\Leftrightarrow\left(x-4-3\right)\left(x-4+3\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)
Vậy...
(x-4)-9=0
⇔ (x-4-3)(x-4+3)=0
⇔( x-7)(x-1)=0
⇔\(\left[{}\begin{matrix}x-7=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)
vậy x=7;x=1
co 2 cach de lam bai nay c1 nhu 2 ban kia c2 la nhu nay: