answer:Given the equation: [ (x-y-1)^3 + (y-z-2)^3 + (z-x+3)^3 = 18, ] we introduce the variables (a), (b), and (c) such that: [ a = x - y - 1, quad b = y - z - 2, quad c = z - x + 3. ] Since we have: [ a + b + c = (x - y - 1) + (y - z - 2) + (z - x + 3), ] we can simplify this: [ a + b + c = x - y - 1 + y - z - 2 + z - x + 3 = 0. ] Thus, [ a + b + c = 0. ] Next, we use the identity for the sum of cubes of three numbers that add up to zero: [ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) = 0. ] Since (a + b + c = 0), it follows that: [ a^3 + b^3 + c^3 = 3abc. ] Given in the problem, (a^3 + b^3 + c^3 = 18), we substitute: [ 3abc = 18, ] implying: [ abc = 6. ] Now, we find the integer solutions for ((a, b, c)) where (a + b + c = 0) and (abc = 6). The possible integer triplets ((a, b, c)) that satisfy these conditions are: [ (3, -2, -1), , (3, -1, -2), , (1, 2, -3), , (-1, -2, 3), , (-1, 3, -2), , (-2, 3, -1). ] We evaluate each of these triplets to determine the solutions for (x, y,) and (z): 1. **Triplet ((a, b, c) = (-1, -2, 3))**: - (x - y - 1 = -1 implies x = y), - (y - z - 2 = -2 implies y = z), - (z - x + 3 = 3). This yields: [ x = y = z. ] 2. **Triplet ((a, b, c) = (-2, -1, 3))**: - (x - y - 1 = -2 implies x = y - 1), - (y - z - 2 = -1 implies y = z + 1), - (z - x + 3 = 3). This yields: [ x = z = y - 1. ] 3. **Triplet ((a, b, c) = (-1, 3, -2))**: - (x - y - 1 = -1 implies x = y), - (y - z - 2 = 3 implies y = z + 5), - (z - x + 3 = -2 implies z = x - 1). This yields: [ x = y = z + 5. ] 4. **Triplet ((a, b, c) = (3, -1, -2))**: - (x - y - 1 = 3 implies x = y + 4), - (y - z - 2 = -1 implies y = z + 1), - (z - x + 3 = -2 implies z = x - 5). This yields: [ x = y + 4 = z + 5. ] 5. **Triplet ((a, b, c) = (3, -2, -1))**: - (x - y - 1 = 3 implies x = y + 4), - (y - z - 2 = -2 implies y = z + 4), - (z - x + 3 = -1 implies z = x - 4). This yields: [ x = y + 4 = z + 4. ] 6. **Triplet ((a, b, c) = (-2, 3, -1))**: - (x - y - 1 = -2 implies x = y - 1), - (y - z - 2 = 3 implies y = z + 5), - (z - x + 3 = -1 implies z = x - 4). This yields: [ x = y - 1 = z + 4. ] Thus, the solutions for ((x, y, z)) are: [ begin{gathered} x = y = z, quad x = y - 1 = z, x = y = z + 5, quad x = y + 4 = z + 5, x = y + 4 = z + 4, quad x = y - 1 = z + 4, end{gathered} ] where (x, y,) and (z) are integers. (boxed{text{This completes the solution.}})

answer:We are given the medians (m_a), (m_b), and (m_c) of triangle (ABC), and we need to find its area (S_{ABC}). 1. **Understanding the Relationship Between Triangle Areas:** It is a known fact that if we construct a triangle using the medians of any triangle, the area of this new triangle (let's call it ( triangle DEF )) is (frac{3}{4}) of the area of the original triangle ( triangle ABC ). Therefore, [ S_m = frac{3}{4} S_{ABC} ] where (S_m) is the area of the triangle formed by the medians. 2. **Applying Heron's Formula to Find (S_m):** Using Heron's formula, we can find the area (S_m) of the triangle whose sides are the medians (m_a), (m_b), and (m_c). Let's recall Heron's formula for area: [ S = sqrt{s(s-a)(s-b)(s-c)} ] where (s) is the semi-perimeter of the triangle: [ s = frac{m_a + m_b + m_c}{2} ] Hence, the area (S_m) is given by: [ S_m = sqrt{s(s - m_a)(s - m_b)(s - m_c)} ] 3. **Relating (S_m) to (S_{ABC}):** From step 1, we have: [ S_m = frac{3}{4} S_{ABC} ] 4. **Solving for (S_{ABC}):** Rearranging the relation from step 3, we obtain: [ S_{ABC} = frac{4}{3} S_m ] 5. **Final Calculation:** After calculating (S_m) using Heron's formula, we substitute it back into the equation to get (S_{ABC}). In conclusion, the area of triangle (ABC) is given by: [ boxed{S_{ABC} = frac{4}{3} sqrt{left(frac{m_a + m_b + m_c}{2}right)left(frac{m_a + m_b + m_c}{2} - m_aright)left(frac{m_a + m_b + m_c}{2} - m_bright)left(frac{m_a + m_b + m_c}{2} - m_cright)}} ]

answer:Since the universal set U={1,2,3,4,5,6}, and the set A={2,3,6}, the complement of set A in U is {1,4,5}, therefore, the correct answer is boxed{C}.

answer:1. **Step 1: Analyze the constraints** 1.1. Three and Five dislike the toy fish and Bohemia ball: [ text{Three:} quad text{No to toy fish and Bohemia ball} text{Five:} quad text{No to teaser stick and Bohemia ball} ] 1.2. Exactly one between One and Two gets the toy fish, and the other gets the treadmill. [ text{One or Two:} quad text{One gets toy fish, and other gets treadmill} ] 1.3. If Four doesn’t get the teaser stick or the treadmill, Four borrows these toys from other cats: [ text{Four:} quad text{Borrows teaser stick or treadmill otherwise} ] 1.4. Two and Three do not share their toys: [ text{Two and Three:}quad text{They don't share toys} ] 2. **Step 2: Determine the toy for Three (三猫)** 2.1. Three dislikes the toy fish and Bohemia ball, so potential toys for Three are the hairball, teaser stick, and treadmill. 2.2. As a special condition, Two and Three do not share toys. Assuming Two has either the toy fish or treadmill, Three must have the hairball. [ ∴ text{Three gets the hairball} ] 3. **Step 3: Determine toy for Five (五猫)** 3.1. Five dislikes the teaser stick and Bohemia ball, thus potential toys for Five are the hairball, treadmill, and toy fish. 3.2. Since the hairball is taken by Three, Five can only have the treadmill or toy fish. [ ∴ text{Five gets the treadmill or toy fish} ] 4. **Step 4: Determine the toys for One and Two (大猫 and 二猫)** 4.1. Either One or Two gets the toy fish and the other gets the treadmill according to given constraint (1). 4.2. Three has the toy fish. Assume Five has the treadmill. Check possible placements: - If One gets toy fish, then Two gets treadmill. - This is valid and respects that Three gets a separate toy. 5. **Step 5: Determine toy for Four (四猫)** 5.1. Given Four borrows any non-allocated toy, with previous assumptions: [ ∴ text{Four borrow or receive any toy except teaser stick or treadmill} ] 6. **Conclusion: What toy does One get？** 6.1. Since Five does not get the treadmill, the only valid constraint assignment result is One (大猫) getting the treadmill. [ boxed{B} ]