answer:Given that m is a root of the equation x^{2}+3x-2022=0, we proceed as follows: 1. Since m is a root, we substitute m into the equation: [m^{2}+3m-2022=0] 2. We can express m^{2} in terms of m: [m^{2}=-3m+2022] 3. To find m^{3}, we multiply both sides of the equation by m: [m^{3}=m(-3m+2022)=-3m^{2}+2022m] 4. Substituting m^{2}=-3m+2022 into the expression for m^{3}: [m^{3}=-3(-3m+2022)+2022m=2031m-6066] 5. Now, we calculate m^{3}+4m^{2}-2019m-2023 using the expressions for m^{2} and m^{3}: begin{align*} m^{3}+4m^{2}-2019m-2023 &= 2031m-6066+4(-3m+2022)-2019m-2023 &= 2031m-6066-12m+8088-2019m-2023 &= -1 end{align*} Hence, the value of m^{3}+4m^{2}-2019m-2023 is boxed{-1}.

answer:We are given that (n), (m), and (k) are natural numbers and (m geqslant n). We need to prove that if 1 + 2 + cdots + n = mk, then it is possible to partition the numbers (1, 2, cdots, n) into (k) groups such that the sum of the numbers in each group is (m). 1. **Base Case: (n = 1)** When (n = 1), we have (1 = m). The conclusion is trivially true since the single number can form one group of sum (m). 2. **Inductive Hypothesis:** Assume the statement is true for all values less than (n). That is, if the summation of some lesser number of terms (1 + 2 + cdots + n') equals (mk') for (n' < n), then the numbers can be partitioned into (k') groups each having sum (m). 3. **Considering (S_n = {1, 2, cdots, n})** We need to analyze the set (S_n = {1, 2, cdots, n}) under the conditions given in the problem. - **Case (m = n):** If (m = n), then ( frac{1}{2}(n+1) = k ) is an integer. This implies that the total sum is perfectly divisible. We can partition the numbers as follows: [ {{n}, {1, n-1}, {2, n-2}, cdots, {frac{n-1}{2}, frac{n+1}{2}}}. ] - **Case (m = n + 1):** If (m = n + 1), then (n) must be even. We can partition the numbers as: [ {{1, n}, {2, n-1}, cdots, {frac{n}{2}, frac{n}{2} + 1}} ] 4. **Divide into Three Conditions for General (m):** - **Condition 1: (n + 1 < m < 2n) and (m) is odd** Extract a subset (S_{m-n-1} = {1, 2, cdots, m-n-1}). The remaining numbers can be paired such that each pair sums to (m): [ {{m-n, n}, {m-n+1, n-1}, cdots, {frac{m-1}{2}, frac{m+1}{2}}} ] For the subset (S_{m-n-1}), the sum can be expressed as: [ sum S_{m-n-1} = frac{1}{2}(m-n-1)(m-n) ] Simplify it to: [ frac{1}{2} (m-n-1)(m-n) = frac{1}{2}(m^2 - m(2n+1)) + frac{1}{2}n(n+1) ] This sum is divisible by (m). - **Condition 2: (n + 1 < m < 2n) and (m) is even** Extract the subset (S_{m-n-1}), reconstruct the remaining sums such as above for odd (m), and consider the middle number (frac{m}{2}). - **Condition 3: (m geqslant 2n)** In this case: [ k = frac{n(n+1)}{2m} leq frac{1}{4}(n+1) ] Hence, [ n - 2k geq 2k - 1 > 0 ] Consider (S_{n-2k}). We calculate the sum: [ S_{n-2k} = frac{1}{2}(n - 2k)(n - 2k + 1) = frac{1}{2}n(n+1) - k(2n + 1) + 2k^2 ] This sum is divisible by (k). Remaining numbers can be paired: [ {{n - 2k + 1, n}, {n - 2k + 2, n-1}, cdots} ] 5. **Combining Results:** Combining the partitions based on the inductive step shows the given statement holds for the general (n) and (m). Hence, for any (n) that satisfies the conditions, we can always partition the sequence into (k) groups with equal sum (m). Thus, the proof is complete. The given partitions ensure each set sums to (m). (blacksquare)

answer:By the definition of an ellipse, we have (|PF_1| + |PF_2| = 2a = 4), which gives us a = 2. Substituting point P(1, frac{3}{2}) into the equation of the ellipse, we get frac{1}{a^2} + frac{9}{4b^2} = 1, solving for b yields b = sqrt{3}. Thus, the equation of the ellipse is boxed{frac{x^2}{4} + frac{y^2}{3} = 1}. Using the definition of an ellipse, we find a = 2, and then using the fact that point P satisfies the equation of the ellipse, we solve for b and obtain the equation of the ellipse. This problem tests the ability to find the equation of an ellipse, focusing on the application of the definition of an ellipse and point satisfaction. It primarily assesses computational skills and is considered a basic problem.

answer:Given: - Points (X, Y,) and (Z) are the midpoints of segments (AB, AD,) and (BC) respectively. - (XY) is perpendicular to (AB) and (YZ) is perpendicular to (BC). - (angle ABC = 100^circ). - We need to find the measure of (angle ACD). **Step 1: Define angles in isosceles triangles** - Consider triangles (AYB), (BYC), (AYC), and (CYD). These triangles are isosceles since (X, Y, Z) are midpoints. The equal angles are: [ angle BAY = angle ABY = x, quad angle CBY = angle BCY = y, quad angle DCY = angle CDY = z, quad angle CAY = angle ACY = v ] - Given: [ x + y = 100^circ ] **Step 2: Sum of angles in quadrilateral (ABCD)** - Sum of internal angles in any quadrilateral equals (360^circ): [ x + x + y + y + z + z = 360^circ ] Simplifies to: [ 2x + 2y + 2z = 360^circ implies x + y + z = 180^circ ] Substitute (x + y = 100^circ): [ 100^circ + z = 180^circ implies z = 80^circ ] **Step 3: Calculate (angle C Y D) in triangle (CYD)** - Since (triangle CYD) is isosceles with (angle DCY = angle CDY = z = 80^circ): [ angle CYD = 180^circ - 2 cdot 80^circ = 20^circ ] **Step 4: Determine (v)** - (angle CYD) is an external angle for (triangle AYC): [ angle CYD = 2v implies 20^circ = 2v implies v = 10^circ ] **Step 5: Compute (angle ACD)** - The desired angle is: [ angle ACD = z + v = 80^circ + 10^circ = 90^circ ] Conclusion: [ boxed{90^circ} ]