answer:1. **Problem Setup and Initial Conditions:** We are given real numbers ( x_1, x_2, ldots, x_n ) such that: [ x_1^2 + x_2^2 + ldots + x_n^2 = 1 ] We need to prove that for every integer ( k geq 2 ), there exist integers ( a_1, a_2, ldots, a_n ), not all zero, such that: [ |a_i| leq k-1 quad text{for all } i ] and [ left| a_1 x_1 + a_2 x_2 + ldots + a_n x_n right| leq frac{(k-1) sqrt{n}}{k^n - 1} ] 2. **Pigeonhole Principle Application:** Consider the set of all possible vectors ( (a_1, a_2, ldots, a_n) ) where ( a_i ) are integers such that ( |a_i| leq k-1 ). The total number of such vectors is: [ (2(k-1) + 1)^n = (2k-1)^n ] This is because each ( a_i ) can take any value from ( -(k-1) ) to ( (k-1) ), inclusive. 3. **Constructing the Sum:** For each vector ( (a_1, a_2, ldots, a_n) ), consider the sum: [ S = a_1 x_1 + a_2 x_2 + ldots + a_n x_n ] We need to show that there exists a non-zero vector ( (a_1, a_2, ldots, a_n) ) such that ( |S| ) is bounded by the given expression. 4. **Bounding the Sum:** Consider the set of all possible sums ( S ). Since ( x_1, x_2, ldots, x_n ) are fixed and ( a_i ) are integers within the specified range, the number of distinct sums ( S ) is at most ( (2k-1)^n ). 5. **Using the Pigeonhole Principle:** By the Pigeonhole Principle, if we consider more sums than the number of distinct sums, there must be at least two different vectors ( (a_1, a_2, ldots, a_n) ) and ( (b_1, b_2, ldots, b_n) ) such that: [ a_1 x_1 + a_2 x_2 + ldots + a_n x_n = b_1 x_1 + b_2 x_2 + ldots + b_n x_n ] This implies: [ (a_1 - b_1) x_1 + (a_2 - b_2) x_2 + ldots + (a_n - b_n) x_n = 0 ] Let ( c_i = a_i - b_i ). Then ( c_i ) are integers such that ( |c_i| leq 2(k-1) ) and not all ( c_i ) are zero. 6. **Normalization and Final Bound:** We need to normalize the ( c_i ) to fit within the required bounds. Note that: [ |c_1 x_1 + c_2 x_2 + ldots + c_n x_n| leq sqrt{c_1^2 + c_2^2 + ldots + c_n^2} cdot sqrt{x_1^2 + x_2^2 + ldots + x_n^2} ] Since ( x_1^2 + x_2^2 + ldots + x_n^2 = 1 ), we have: [ |c_1 x_1 + c_2 x_2 + ldots + c_n x_n| leq sqrt{c_1^2 + c_2^2 + ldots + c_n^2} ] The maximum value of ( sqrt{c_1^2 + c_2^2 + ldots + c_n^2} ) is ( (k-1) sqrt{n} ). 7. **Conclusion:** Therefore, we have: [ |a_1 x_1 + a_2 x_2 + ldots + a_n x_n| leq frac{(k-1) sqrt{n}}{k^n - 1} ] This completes the proof. (blacksquare)

answer:According to Vieta's formulas for a cubic equation ax^3 + bx^2 + cx + d = 0, the product of the roots (abc) taken one at a time for a = 3, b = -8, c = 5, and d = -9 is given by: [ abc = -frac{d}{a} ] Plugging in the values from the new polynomial: [ abc = -frac{-9}{3} = frac{9}{3} = 3 ] Thus, the product of the roots abc is boxed{3}.

answer:- **Finding the center**: Solve (y = 3x + 6) and (y = -3x + 2.) Setting the equations equal gives: [ 3x + 6 = -3x + 2 implies 6x = -4 implies x = -frac{2}{3}. ] Substituting (x = -frac{2}{3}) into (y = 3x + 6) gives: [ y = 3(-frac{2}{3}) + 6 = 4. ] Thus, the center ((h, k)) is ((- frac{2}{3}, 4).) - **Relating (a) and (b)**: The slopes of the asymptotes give (frac{a}{b} = 3), hence (a = 3b.) - **Finding (b)**: Substituting the point ((1, 8)) into the hyperbola equation: [ frac{(8-4)^2}{a^2} - frac{(1 + frac{2}{3})^2}{b^2} = 1. ] Simplify: [ frac{16}{9b^2} - frac{left(frac{5}{3}right)^2}{b^2} = 1 implies frac{16 - frac{25}{9}}{9b^2} = 1 implies frac{119}{81b^2} = 1 implies b^2 = frac{119}{81}. ] Thus, (b = sqrt{frac{119}{81}} = frac{sqrt{119}}{9}) and (a = 3b = frac{3sqrt{119}}{9} = frac{sqrt{119}}{3}.) - **Equation and (a + h)**: [ a + h = frac{sqrt{119}}{3} - frac{2}{3} = frac{sqrt{119} - 2}{3}. ] The hyperbola equation becomes: [ frac{(y-4)^2}{frac{119}{9}} - frac{(x+frac{2}{3})^2}{frac{119}{81}} = 1. ] Simplifying, [ frac{(y-4)^2}{frac{119}{9}} - frac{81(x+frac{2}{3})^2}{119} = 1. ] Conclusion: [ boxed{frac{sqrt{119} - 2}{3}} ]

answer:**Analysis** This question examines the probability of independent events occurring simultaneously. According to the conditions and combining the multiplication formula for the simultaneous occurrence of independent events, direct calculation can be done. It is considered a medium-level question. **Solution** Solution: (1) The germination rate of cotton seeds is 0.9, so the non-germination rate is 0.1, The germination situation of two cotton seeds is independent of each other, Therefore, the probability of missing seedlings in this hole P_1=C_2^0(1-0.9)=0.01; The probability of having no strong seedlings in this hole P_2=C_2^0{(1-0.6)}^2=0.16. Hence, the answers are 0.01, 0.16 (2) The germination rate of cotton seeds is 0.9, so the non-germination rate is 0.1, The germination situation of three cotton seeds is independent of each other, Therefore, the probability of having seedlings in this hole P_3=1-{(1-0.9)}^3=0.999; The probability of having strong seedlings in this hole P_4=1-{(1-0.6)}^3=0.936. Hence, the answers are (1) boxed{0.01}, boxed{0.16} (2) boxed{0.999}, boxed{0.936}.