answer:1. **Assumption:** - Assume that the quadratic polynomials ( f(x) = x^2 + mx + n ) and ( p(x) = x^2 + kx + l ) have a common root ( x_0 ). 2. **Observation:** - Given that ( k > m > n > l > 0 ), all coefficients of both polynomials are positive. 3. **Nature of Roots:** - Since the coefficients are all positive, any real roots the polynomials might have must be negative. Hence, ( x_0 < 0 ). 4. **Equation for the Common Root:** - Since ( x_0 ) is a common root of both polynomials, it satisfies: begin{align*} f(x_0) &= x_0^2 + mx_0 + n = 0, p(x_0) &= x_0^2 + kx_0 + l = 0. end{align*} 5. **Forming the Equality:** - Subtract the second equation from the first to eliminate ( x_0^2 ): begin{align*} (x_0^2 + mx_0 + n) - (x_0^2 + kx_0 + l) &= 0, mx_0 + n - kx_0 - l &= 0, (m - k)x_0 + (n - l) &= 0. end{align*} 6. **Solving for ( x_0 ):** - Solve for ( x_0 ): begin{align*} (m - k)x_0 &= l - n, x_0 &= frac{l - n}{m - k}. end{align*} 7. **Analysis of Fractions:** - Given ( k > m ), we have ( m - k < 0 ). - Similarly, since ( n > l ), we get ( l - n < 0 ). 8. **Sign of ( x_0 ):** - The term ( frac{l - n}{m - k} ) is a ratio of two negative numbers, so it is positive. Hence: [ x_0 > 0. ] 9. **Contradiction:** - This gives a contradiction because we initially concluded that ( x_0 < 0 ). # Conclusion: Therefore, the initial assumption is incorrect; the polynomials ( f(x) ) and ( p(x) ) cannot have a common root. [ boxed{text{No}} ]

answer:Solution: (1) From begin{cases} m^2 - 2m = 0 m^2 + m - 6 neq 0 end{cases}, we solve and get m = 0. Therefore, if the point corresponding to the complex number z = (m^2 - 2m) + (m^2 + m - 6)i is on the imaginary axis, then m = 0; (2) For the point corresponding to the complex number z = (m^2 - 2m) + (m^2 + m - 6)i to be in the third quadrant, we have begin{cases} m^2 - 2m < 0 m^2 + m - 6 < 0 end{cases}, solving this gives 0 < m < 2. Therefore, the solutions are m = boxed{0} for the point to be on the imaginary axis, and 0 < m < 2 for the point to be in the third quadrant, which can be written as m in boxed{(0, 2)}.

answer:Simplify the congruence: [ 5(12x + 2) equiv 3 pmod p implies 60x + 10 equiv 3 pmod p implies 60x equiv -7 pmod p implies 60x equiv p - 7 pmod p. ] This equation is solvable if and only if 60 is invertible modulo p. This requires that gcd(60, p) = 1. The prime divisors of 60 are 2, 3, 5. Therefore, if p is any of these primes, then gcd(60, p) > 1 and there is no solution for x. Thus, the primes that make the congruence unsolvable are 2, 3, and 5. Adding these primes gives: [ 2 + 3 + 5 = boxed{10}. ]

answer:# Step-by-Step Solution: Part (1): Comparing the Probability of Winning Between A and B - **Calculating Probability for A:** Let A_{1} be the event where "A wins in the first round" and A_{2} be the event where "A wins in the second round." The probability of A winning both rounds and hence the competition can be calculated as follows: [P(A_{1}A_{2}) = P(A_{1}) times P(A_{2}) = frac{3}{5} times frac{3}{5} = frac{9}{25}.] - **Calculating Probability for B:** Similarly, let B_{1} be the event where "B wins in the first round" and B_{2} be the event where "B wins in the second round." The probability of B winning the competition is: [P(B_{1}B_{2}) = P(B_{1}) times P(B_{2}) = frac{3}{4} times frac{1}{2} = frac{3}{8}.] - **Comparing Probabilities:** Since frac{9}{25} < frac{3}{8}, the probability of B winning the competition is higher. Therefore, B has a higher probability of winning the competition. Hence, the answer to the first question is boxed{B} has a higher probability of winning. Part (2): Probability That At Least One Wins the Competition - **Calculating Complementary Events:** Let C be the event where "A wins the competition" and D be the event where "B wins the competition." The complementary events are P(overline{C}) = 1 - P(A_{1}A_{2}) and P(overline{D}) = 1 - P(B_{1}B_{2}), which gives us: [P(overline{C}) = 1 - frac{9}{25} = frac{16}{25},] [P(overline{D}) = 1 - frac{3}{8} = frac{5}{8}.] - **Calculating Probability of At Least One Winning:** The event "Ccup D" represents "at least one of them wins the competition." The probability that at least one of them wins is: [P(Ccup D) = 1 - P(overline{C}overline{D}) = 1 - P(overline{C})P(overline{D}) = 1 - left(frac{16}{25} times frac{5}{8}right) = frac{3}{5}.] Therefore, the probability that at least one of them wins the competition is boxed{frac{3}{5}}.