Solve the inequation: $\;$ $\left|x^2 - 2x\right| < x$
Given inequation: $\;$ $\left|x^2 - 2x\right| < x$ $\;\;\; \cdots \; (1)$
Case 1:
When $\;$ $x^2 - 2x > 0$
i.e. $\;$ when $\;$ $x \left(x - 2\right) > 0$
i.e. $\;$ when $\;$ $x > 0 \; \cap \; x - 2 > 0$ $\;\;$ OR $\;\;$ $x < 0 \; \cap \; x - 2 < 0$
i.e. $\;$ when $\;$ $x > 0 \; \cap \; x > 2$ $\;\;$ OR $\;\;$ $x < 0 \; \cap \; x < 2$
i.e. $\;$ when $\;$ $x > 2$ $\;$ OR $\;$ $x < 2$, $\;\;$ $\left|x^2 - 2x\right| = x^2 - 2x$
Then $(1)$ becomes
$x^2 - 2x < x$
i.e. $\;$ $x^2 - 3x < 0$
i.e. $\;$ $x \left(x - 3\right) < 0$
i.e. $\;$ $x < 0 \; \cap \; x - 3 > 0$ $\;$ OR $\;$ $x > 0 \; \cap \; x - 3 < 0$
i.e. $\;$ $x < 0 \; \cap \; x > 3$ $\;\;$ which is not possible
OR $\;$ $x > 0 \; \cap \; x < 3$ $\implies$ $x \in \left(0, 3\right)$
$\therefore \;$ When $\;$ $x^2 - 2x > 0$, $\;$ $x \in \left(0, 3\right)$ $\;\;\; \cdots \; (2)$
Case 2:
When $\;$ $x^2 - 2x < 0$
i.e. $\;$ when $\;$ $x \left(x - 2\right) < 0$
i.e. $\;$ when $\;$ $x < 0 \; \cap \; x - 2 > 0$ $\;\;$ OR $\;\;$ $x > 0 \; \cap \; x - 2 < 0$
i.e. $\;$ when $\;$ $x < 0 \; \cap \; x > 2$ $\;\;$ which is not possible
OR $\;\;$ $x > 0 \; \cap \; x < 2$ $\implies$ $x \in \left(0,2\right)$
i.e. $\;$ when $\;$ $x \in \left(0, 2\right)$, $\;\;$ $\left|x^2 - 2x\right| = -x^2 + 2x$
Then $(1)$ becomes
$-x^2 + 2x < x$
i.e. $\;$ $-x^2 + x < 0$
i.e. $\;$ $x^2 - x > 0$
i.e. $\;$ $x \left(x - 1\right) > 0$
i.e. $\;$ $x < 0 \; \cap \; x - 1 < 0$ $\;$ OR $\;$ $x > 0 \; \cap \; x - 1 > 0$
i.e. $\;$ $x < 0 \; \cap \; x < 1$ $\;$ OR $\;$ $x > 0 \; \cap \; x > 1$
i.e. $\;$ $x < 0$ $\;$ OR $\;$ $x > 1$ $\implies$ $x \in \left(x < 0 \; \cup \; x > 1\right)$
$\therefore \;$ When $\;$ $x^2 - 2x < 0$, $\;$ $x \in \left(x < 0 \; \cup \; x > 1\right)$ $\;\;\; \cdots \; (3)$
$\therefore \;$ In view of $(2)$ and $(3)$, solution of $(1)$ is $\;\;$ $x \in \left(1, 3\right)$