Найдите область определения функции:y=log2(25-x^2)+1/(3x-4)

Решите задачу:

$y=log_2(25-x^2)+ \frac{1}{3x-4} \\\\ODZ:\; \; \left \{ {{25-x^2\ \textgreater \ 0} \atop {3x-4\ne 0}} \right. \; \; \left \{ {{x^2-25\ \textless \ 0} \atop {x\ne \frac{4}{3}}} \right. \; \; \left \{ {{(x-5)(x+5)\ \textless \ 0} \atop {x\ne 1\frac{1}{3}}} \right. \\\\(x-5)(x+5)\ \textless \ 0\qquad +++(-5)---(5)+++\\\\ \left \{ {{x\in (-5,5)} \atop {x\ne 1\frac{1}{3}}} \right. \qquad -----(-5)///(1\frac{1}{3})///(5)-----\\\\Otvet:\quad x\in (-5,1\frac{1}{3})\cup (1\frac{1}{3},5)$