Step 1: Since $x^2 = \vert{}x\vert{}^2$, rewrite as $\vert{}x\vert{}^2 - 5\vert{}x\vert{} + 6 < 0$.

Step 2: Factor: $(\vert{}x\vert{} - 2)(\vert{}x\vert{} - 3) < 0 \implies 2 < \vert{}x\vert{} < 3$.

Step 3: Solving $2 < \vert{}x\vert{} < 3$ yields two intervals: $2 < x < 3$ or $-3 < x < -2$.

Step 1: Rewrite as $\vert{}x\vert{}^2 - \vert{}x\vert{} - 2 \ge 0$.

Step 2: Factor: $(\vert{}x\vert{} - 2)(\vert{}x\vert{} + 1) \ge 0$.

Step 3: Since $\vert{}x\vert{} + 1 > 0$ for all real $x$, this reduces to $\vert{}x\vert{} - 2 \ge 0 \implies \vert{}x\vert{} \ge 2$.

Step 4: $x \le -2 \text{ or } x \ge 2$.

Answer: (b) x $\le$ –2 or x $\ge$ 2

Step 1: Rewrite $(3 - x)$ as $-(x - 3)$, so: $-(x - 1)(x - 3)(x - 2)^2 > 0 \implies (x - 1)(x - 3)(x - 2)^2 < 0$.

Step 2: Note $(x - 2)^2 > 0$ for all $x \neq 2$. At $x = 2$, the expression equals $0$.

Step 3: Divide by $(x - 2)^2$ for $x \neq 2$: $(x - 1)(x - 3) < 0 \implies 1 < x < 3$.

Step 4: Exclude $x = 2$: $1 < x < 3, x \neq 2$.

Answer: (d) 1 < x < 3, x $\neq$ 2

Step 1: Simplify: $x^2 - x + 1 > 0$.

Step 2: Discriminant $D = (-1)^2 - 4(1)(1) = -3 < 0$. Since $a = 1 > 0$ and $D < 0$, the quadratic expression is always positive for all real $x$.

Answer: (d) For all real x

Step 1: Factor first part: $(x - 2)(x - 3)$.

Step 2: Roots of $x^2 + x - 1 = 0$ are $\frac{-1 \pm \sqrt{5}}{2} \approx -1.618, 0.618$.

Step 3: Analyzing intervals of $(x - 2)(x - 3)(x^2 + x - 1) < 0$ gives negative regions between roots:

Step 1: Since $x^2 + 1 > 0$ for all real $x$, divide by $(x^2 + 1)$.

Step 2: Solve $x^2 + 2x - 3 < 0 \implies (x + 3)(x - 1) < 0$.

Step 3: $-3 < x < 1$.

Step 1: Rewrite $(-1 - x)$ as $-(x + 1)$: $-(x - 1)(x + 1)(x + 2)^2 < 0 \implies (x - 1)(x + 1)(x + 2)^2 > 0$.

Step 2: Since $(x + 2)^2 > 0$ for $x \neq -2$, solve $(x - 1)(x + 1) > 0$.

Step 3: $x < -1 \text{ or } x > 1$, with $x \neq -2$.

Step 1: Factor expressions: $(x + 2)^2 \cdot [-(x - 1)^2] > 0 \implies -(x + 2)^2 (x - 1)^2 > 0$.

Step 2: The left-hand side is non-positive for all real $x$ because squares are non-negative. It can never be strictly positive ($> 0$).

Step 1: Let $y = x^2$: $y^2 - 5y + 4 < 0 \implies (y - 1)(y - 4) < 0 \implies 1 < y < 4$.

Step 2: Substitute back: $1 < x^2 < 4$.

Step 3: Solution set: $-2 < x < -1 \text{ or } 1 < x < 2$.

Answer: (c) –2 < x < –1 or 1 < x < 2

Step 1: Factor: $(x^2 - 9)(x^2 + 7) \le 0$.

Step 2: Since $x^2 + 7 > 0$ for all real $x$, divide by $(x^2 + 7)$: $x^2 - 9 \le 0$.

Step 3: $(x - 3)(x + 3) \le 0 \implies -3 \le x \le 3$.