Алгебра: 1)(x-2)^3=-216 2)log4(4+7x)=log4(1+5x)+1 Решите плиз

$1)(x-2)^3=-216 \\ \\ \sqrt[3]{(x-2)^3} = \sqrt[3]{-216} \\ \\ (x-2) = -6 \\ \\ x = -4$

$2) log_4(4+7x)=log_4(1+5x)+1 \\ \\ log_4(4+7x)=log_4(1+5x)+log_44 \\ \\ log_4(4+7x)=log_44(1+5x) \\ \\ 4+7x = 4(1+5x) \\ \\ 4+7x = 4+20x \\ \\ 4-4 = 20x - 7x \\ \\ 13x = 0 \\ \\ x = 0$

ОДЗ:

$\left \{ {{4+7x \ \textgreater \ 0} \atop {1+5x \ \textgreater \ 0}} \right. \ \ \left \{ {{x \ \textgreater \ - \frac{4}{7} } \atop {x \ \textgreater \ -\frac{1}{5} }} \right. \Rightarrow x \ \textgreater \ -\frac{1}{5}$

$x = 0 \in ODZ$

Ответ: $x = 0$