Решите неравенство 4/x^2-4x меньше 1/x-4

Решите задачу:

$\frac{4}{x^2-4x}\ \textless \ \frac{1}{x-4}\\\frac{4}{x(x-4)} \ \textless \ \frac{1}{x-4}\\\frac{4}{x(x-4)} \ \textless \ \frac{x}{x(x-4)}\\\frac{4-x}{x(x-4)} \ \textless \ 0\\ \\4-x \ \textless \ 0\\x \ \textgreater \ 4;\\\\x(x-4) \ \ \textless \ \ 0\\x_1 \neq 0\\x_2 \neq 4 \\\\x\in (0;4) \cup (4; + \infty)$