1. **Problem 1:** Solve the equation $$\left\lceil 2\lfloor x \rfloor + \Delta + 2\right\rceil = 1$$. - Here, $\lfloor x \rfloor$ is the floor function (greatest integer less than or equal to $x$). - $\Delta$ is not defined in the problem, so we cannot solve this equation without knowing $\Delta$. 2. **Problem 2:** Solve the equation $$|x-1| + |x-2| = 3$$. - The absolute value function $|a|$ measures the distance of $a$ from zero. - We analyze the expression by considering intervals defined by points 1 and 2. **Step 1:** Consider intervals: - For $x \leq 1$, $|x-1| = 1 - x$, $|x-2| = 2 - x$. - For $1 < x < 2$, $|x-1| = x - 1$, $|x-2| = 2 - x$. - For $x \geq 2$, $|x-1| = x - 1$, $|x-2| = x - 2$. **Step 2:** Solve in each interval: - If $x \leq 1$: $|x-1| + |x-2| = (1 - x) + (2 - x) = 3 - 2x = 3 \implies 3 - 2x = 3 \implies 2x = 0 \implies x = 0$. - Check if $x=0 \leq 1$: yes, so $x=0$ is a solution. - If $1 < x < 2$: $|x-1| + |x-2| = (x - 1) + (2 - x) = 1$ which is not equal to 3, so no solution here. - If $x \geq 2$: $|x-1| + |x-2| = (x - 1) + (x - 2) = 2x - 3 = 3 \implies 2x = 6 \implies x = 3$. - Check if $x=3 \geq 2$: yes, so $x=3$ is a solution. **Final solutions:** $x = 0$ and $x = 3$. 3. **Problem 3:** Given sets $$C = \{1, 3\}, \quad B = \{2, 5\}, \quad A = \{1, 2, 3\}$$ - $M = \{1, ..., 1\}$ is not well-defined (all elements are 1?), so we skip it. 4. **Problem 4:** Find the Cartesian product $A \times B$. - The Cartesian product $A \times B$ is the set of all ordered pairs $(a,b)$ where $a \in A$ and $b \in B$. $$A \times B = \{(1,2), (1,5), (2,2), (2,5), (3,2), (3,5)\}$$ 5. **Problem 5:** Find the intersection $(A \times B) \cap C$. - Since $C = \{1, 3\}$ contains numbers, and $A \times B$ contains ordered pairs, their intersection is empty because elements are of different types. **Final answer:** $(A \times B) \cap C = \emptyset$. --- **Summary:** - Problem 1 cannot be solved without definition of $\Delta$. - Problem 2 solutions: $x=0$ and $x=3$. - Problem 4: $A \times B = \{(1,2), (1,5), (2,2), (2,5), (3,2), (3,5)\}$. - Problem 5: $(A \times B) \cap C = \emptyset$.