Решите уравнение:
а) $(x - 3)^2 = 25$;
б) $(x + 4)^2 = 9$;
в) $(x - 6)^2 = 7$;
г) $(x + 2)^2 = 6$.
$(x - 3)^2 = 25$
$x - 3 = ±\sqrt{25}$
x − 3 = ±5
\begin{equation*}
\begin{cases}
x - 3 = 5 &\\
x - 3 = -5 &
\end{cases}
\end{equation*}
\begin{equation*}
\begin{cases}
x = 5 + 3&\\
x = -5 + 3&
\end{cases}
\end{equation*}
\begin{equation*}
\begin{cases}
x = 8&\\
x = -2&
\end{cases}
\end{equation*}
$(x + 4)^2 = 9$
$x + 4 = ±\sqrt{9}$
x + 4 = ±3
\begin{equation*}
\begin{cases}
x + 4 = 3 &\\
x + 4 = -3 &
\end{cases}
\end{equation*}
\begin{equation*}
\begin{cases}
x = 3 - 4&\\
x = -3 - 4&
\end{cases}
\end{equation*}
\begin{equation*}
\begin{cases}
x = -1&\\
x = -7&
\end{cases}
\end{equation*}
$(x - 6)^2 = 7$
$x - 6 = ±\sqrt{7}$
$x = ±\sqrt{7} + 6$
$(x + 2)^2 = 6$
$x + 2 = ±\sqrt{6}$
$x = ±\sqrt{6} - 2$
Пожауйста, оцените решение