1. **Problem 1:** Verify if $2017^2 = 2016^2 + 2016$. 2. Use the identity for the difference of squares: $$a^2 - b^2 = (a-b)(a+b)$$ 3. Calculate $2017^2 - 2016^2$: $$2017^2 - 2016^2 = (2017 - 2016)(2017 + 2016) = 1 \times 4033 = 4033$$ 4. Compare with the right side of the equation: $$2016 \neq 4033$$ 5. Therefore, the equation $2017^2 = 2016^2 + 2016$ is **false**. --- 6. **Problem 2:** Find the value of $xy$ if $[-1, x] \cap [y, 5] = [2, 3]$. 7. The intersection of intervals $[-1, x]$ and $[y, 5]$ is $[2, 3]$. 8. For the intersection to be $[2, 3]$, the overlapping region must start at 2 and end at 3. 9. This means: - The left endpoint of the intersection is the maximum of the two intervals' left endpoints: $\max(-1, y) = 2$ so $y = 2$. - The right endpoint of the intersection is the minimum of the two intervals' right endpoints: $\min(x, 5) = 3$ so $x = 3$. 10. Calculate $xy$: $$xy = 3 \times 2 = 6$$ 11. Since 6 is not among the options (8 or 5), re-check the logic: - The intersection is $[2,3]$. - For $[-1,x]$ to include $[2,3]$, $x \geq 3$. - For $[y,5]$ to include $[2,3]$, $y \leq 2$. 12. The intersection is from $\max(-1,y)$ to $\min(x,5)$, which equals $[2,3]$. 13. So $\max(-1,y) = 2 \Rightarrow y = 2$ (since $y \geq -1$). 14. And $\min(x,5) = 3 \Rightarrow x = 3$ (since $x \leq 5$). 15. Therefore, $xy = 3 \times 2 = 6$. 16. Since 6 is not an option, check if the problem expects $xy = 5$ or $8$. 17. Possibly a typo or misinterpretation; if $y=2$ and $x=4$, intersection is $[2,4]$ which is not $[2,3]$. 18. So the correct $xy$ is 6, but since options are 8 or 5, none match exactly. --- 19. **Problem 3:** When the side length of a square increases, find the percentage increase in area. 20. Let the original side length be $s$. 21. New side length is $s + \Delta s$. 22. Original area: $A = s^2$. 23. New area: $A' = (s + \Delta s)^2 = s^2 + 2s\Delta s + (\Delta s)^2$. 24. Percentage increase in area: $$\frac{A' - A}{A} \times 100 = \frac{2s\Delta s + (\Delta s)^2}{s^2} \times 100 = \left(2 \frac{\Delta s}{s} + \left(\frac{\Delta s}{s}\right)^2\right) \times 100$$ 25. For small increases, $(\frac{\Delta s}{s})^2$ is negligible, so approximate percentage increase is: $$2 \times \frac{\Delta s}{s} \times 100$$ 26. Without a specific increase value, the problem cannot be solved exactly. --- **Final answers:** - Problem 1: False, $2017^2 \neq 2016^2 + 2016$. - Problem 2: $xy = 6$ (not matching given options). - Problem 3: Percentage increase in area is approximately $2 \times$ the percentage increase in side length.