x - 3" alt="\sqrt{x^2 - 4x} > x - 3" align="absmiddle" class="latex-formula">
ОДЗ: ![x^2 - 4x \geq 0 \leftrightarrow x(x - 4) \geq 0 \leftrightarrow x \in [-\infty; 0] \cup [4; \infty]](https://tex.z-dn.net/?f=x%5E2%20-%204x%20%5Cgeq%200%20%5Cleftrightarrow%20x%28x%20-%204%29%20%5Cgeq%200%20%5Cleftrightarrow%20x%20%5Cin%20%5B-%5Cinfty%3B%200%5D%20%5Ccup%20%5B4%3B%20%5Cinfty%5D "x^2 - 4x \geq 0 \leftrightarrow x(x - 4) \geq 0 \leftrightarrow x \in [-\infty; 0] \cup [4; \infty]")
При 
неравенство, очевидно, выполняется.
x - 3 | ^2" alt="\sqrt{x^2 - 4x} > x - 3 | ^2" align="absmiddle" class="latex-formula">
x^2 - 6x + 9" alt="x^2 - 4x > x^2 - 6x + 9" align="absmiddle" class="latex-formula">
9 \rightarrow x > \frac{9}{2}" alt="2x > 9 \rightarrow x > \frac{9}{2}" align="absmiddle" class="latex-formula">
Ответ: ![x \in [-\infty; 0] \cup [\frac{9}{2}; \infty]](https://tex.z-dn.net/?f=x%20%5Cin%20%5B-%5Cinfty%3B%200%5D%20%5Ccup%20%5B%5Cfrac%7B9%7D%7B2%7D%3B%20%5Cinfty%5D "x \in [-\infty; 0] \cup [\frac{9}{2}; \infty]")