answer:To solve the mystery behind Xiao Ming's ability to guess the numbers correctly, let's denote the three positive integers Xiao Yong selects as a, b, and c. Following the steps Xiao Ming described, we can represent the process mathematically: 1. Multiply the first number by 2 and then add 2: 2a + 2. 2. Multiply the result of the first step by 5 and then add the second number: (2a + 2) times 5 + b. 3. Multiply the result of the second step by 10 and then add the third number: ((2a + 2) times 5 + b) times 10 + c. Simplifying this expression, we get: begin{align*} ((2a + 2) times 5 + b) times 10 + c &= (10a + 10 + b) times 10 + c &= 100a + 10b + c + 100. end{align*} To find the original numbers a, b, and c from the final result, we subtract 100 from the final result. This subtraction isolates 100a + 10b + c, where the hundreds place represents a, the tens place represents b, and the units place represents c. Given the "final result" is 567, we calculate: begin{align*} 567 - 100 &= 467. end{align*} From 467, we can deduce: - The hundreds place (4) represents a. - The tens place (6) represents b. - The units place (7) represents c. Therefore, the three positive integers initially selected by Xiao Yong are 4, 6, and 7. Thus, the correct answer is option boxed{text{B: } 4, 6, 7}.

answer:1. Given the system of equations: [ begin{cases} x^2 = 2 + y y^2 = 2 + z z^2 = 2 + x end{cases} ] Let's number the equations for convenience: [ begin{cases} x^2 = 2 + y quad text{(1)} y^2 = 2 + z quad text{(2)} z^2 = 2 + x quad text{(3)} end{cases} ] 2. Subtracting the equations pairwise: [ begin{cases} y^2 - x^2 = z - y quad text{(2) - (1)} z^2 - y^2 = x - z quad text{(3) - (2)} x^2 - z^2 = y - x quad text{(1) - (3)} end{cases} ] Using the difference of squares, we get: [ begin{cases} (y - x)(y + x) = z - y quad text{(4)} (z - y)(z + y) = x - z quad text{(5)} (x - z)(x + z) = y - x quad text{(6)} end{cases} ] 3. Multiplying equations (4), (5), and (6): [ (y - x)(y + x) cdot (z - y)(z + y) cdot (x - z)(x + z) = (z - y)(x - z)(y - x) ] Simplifying, we get: [ (x + y)(y + z)(z + x) = 1 quad text{(7)} ] 4. Rewriting the original equations: [ begin{cases} (x - 1)(x + 1) = y + 1 quad text{(8)} (y - 1)(y + 1) = z + 1 quad text{(9)} (z - 1)(z + 1) = x + 1 quad text{(10)} end{cases} ] Since (x, y, z) are distinct, none of them can be (-1). 5. Multiplying equations (8), (9), and (10): [ (x - 1)(y - 1)(z - 1) = 1 quad text{(11)} ] 6. Adding the original equations: [ x^2 + y^2 + z^2 = 6 + (x + y + z) ] Using the identity ((x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)), we get: [ (x + y + z)^2 - 2(xy + yz + zx) = 6 + (x + y + z) ] Let (a = x + y + z), (b = xy + yz + zx), and (c = xyz): [ a^2 - 2b = 6 + a quad text{(12)} ] 7. Using equations (7), (11), and (12): [ begin{cases} ab - c = 1 quad text{(13)} a - b + c = 2 quad text{(14)} a^2 - a - 2b = 6 quad text{(15)} end{cases} ] 8. Solving for (b) and (c): [ 2 times (13) + 2 times (14) implies 2ab + 2a - 2b = 6 implies 2b(a - 1) + 2a = 6 quad text{(16)} ] From (15): [ 2b = a^2 - a - 6 quad text{(17)} ] Substituting (17) into (16): [ (a^2 - a - 6)(a - 1) + 2a = 6 implies (a + 1)a(a - 3) = 0 ] Thus, (a = -1, 0, 3). 9. Using the values of (a): [ begin{cases} (a, b, c) = (-1, -2, 1) (a, b, c) = (0, -3, -1) (a, b, c) = (3, 0, -1) end{cases} ] 10. Calculating (x^2 + y^2 + z^2): [ x^2 + y^2 + z^2 = a^2 - 2b ] For each case: [ begin{cases} a = -1 implies (-1)^2 - 2(-2) = 1 + 4 = 5 a = 0 implies 0^2 - 2(-3) = 0 + 6 = 6 a = 3 implies 3^2 - 2(0) = 9 end{cases} ] The final answer is (boxed{5, 6, 9}).

answer:To find the range of (a), we analyze propositions (P) and (Q) separately. 1. For proposition (P) to always hold, consider the quadratic (ax^2 + ax + 1). There are two cases: - If (a=0), the inequality simplifies to (1 > 0), which is always true. - If (a neq 0), for the quadratic to be strictly positive for all (x), the discriminant must be negative, and the leading coefficient (a) must be positive. This leads to the following constraint: [ a > 0 quad text{and} quad a^2 - 4a < 0 implies 0 < a < 4.] Combining these, (P) is true for (0 leq a < 4). 2. Proposition (Q) represents a hyperbola if and only if the coefficients of (x^2) and (y^2) have opposite signs, which translates to (a(a-3) < 0 implies 0 < a < 3). Next, consider the combination of (P) or (Q) being true ((P lor Q)), and (P) and (Q) together being false ((P land Q)). This means one of them must be true and the other false. - If (P) is true and (Q) is false, we get (a=0) or (3 leq a < 4). - If (P) is false and (Q) is true, there would be no solution since (a) cannot satisfy (a geq 4) or (a < 0), and (0 < a < 3) simultaneously. Combining everything, the range of (a) is either (a=0) or (3 leq a < 4). Therefore, the final answer is: [ boxed{a=0 text{ or } 3 leq a < 4}. ]

answer:Solution: According to the parametric equation of line l, begin{cases}x=2+2t y=1-tend{cases} (t is the parameter), we get its standard equation as x+2y=4. Let P(2cos theta,sin theta), therefore The distance d from P to l is d= dfrac{|2cos theta+2sin theta-4|}{sqrt{5}} = dfrac{|2sqrt{2}sin (theta+ dfrac{pi}{4})-4|}{sqrt{5}} geqslant dfrac{|2sqrt{2}-4|}{sqrt{5}}= dfrac{4-2sqrt{2}}{sqrt{5}}, Equality holds if and only if sin (theta+ dfrac{pi}{4})=1, that is, theta=2kpi+ dfrac{pi}{4}. At this time, sin theta=cos theta= dfrac{sqrt{2}}{2}, therefore P(sqrt{2}, dfrac{sqrt{2}}{2}). Thus, the point P on ellipse C that minimizes the distance to line l is boxed{P(sqrt{2}, dfrac{sqrt{2}}{2})}.