answer:Given the conditions, let's set A at (-a, 0) and B at (a, 0), and let P be (x, y). From the given condition |PA|^2 - |PB|^2 = m, we can derive: (x+a)^2 + y^2 - (x-a)^2 - y^2 = m, which simplifies to 4ax = m. Therefore, the trajectory equation of P is a line. Hence, the correct choice is boxed{text{A}}.

answer:To find the total number of games played, we can use the formula for combinations since each game is a unique combination of two teams: Number of games = C(n, k) where n is the total number of teams and k is the number of teams in each game (which is 2 since two teams play against each other in each game). So, for 9 teams: Number of games = C(9, 2) The formula for combinations is: C(n, k) = n! / (k! * (n - k)!) Plugging in the values: C(9, 2) = 9! / (2! * (9 - 2)!) C(9, 2) = 9! / (2! * 7!) C(9, 2) = (9 * 8 * 7!) / (2 * 1 * 7!) C(9, 2) = (9 * 8) / (2 * 1) C(9, 2) = 72 / 2 C(9, 2) = 36 So, the total number of games played is boxed{36} .

answer:Given that a, b, c are positive real numbers such that a+b+c leqslant 3. We need to prove two parts: 1. [ 3 > frac{1}{a+1} + frac{1}{b+1} + frac{1}{c+1} geqslant frac{3}{2} ] 2. [ frac{a+1}{a(a+2)} + frac{b+1}{b(b+2)} + frac{c+1}{c(c+2)} geqslant 2 ] Part (1): 1. Since (a, b, c) are positive, we know (frac{1}{a+1} < 1), (frac{1}{b+1} < 1), and (frac{1}{c+1} < 1). 2. Summing them up, we get: [ frac{1}{a+1} + frac{1}{b+1} + frac{1}{c+1} < 3 ] 3. To prove the lower bound, we use the Cauchy-Schwarz inequality in the following way: [ left( frac{1}{a+1} + frac{1}{b+1} + frac{1}{c+1} right) left( (a+1) + (b+1) + (c+1) right) geqslant 9 ] 4. Given (a + b + c leqslant 3), thus: [ (a+1) + (b+1) + (c+1) leqslant 3 + 3 = 6 ] 5. Therefore, [ frac{1}{a+1} + frac{1}{b+1} + frac{1}{c+1} geqslant frac{9}{(a+1) + (b+1) + (c+1)} geqslant frac{9}{6} = frac{3}{2} ] 6. Combining both results, we conclude: [ frac{3}{2} leqslant frac{1}{a+1} + frac{1}{b+1} + frac{1}{c+1} < 3 ] # Conclusion: [ boxed{frac{3}{2} leqslant frac{1}{a+1} + frac{1}{b+1} + frac{1}{c+1} < 3} ] Part (2): 1. Using the given inequality from Part (1) and the Cauchy-Schwarz inequality, we start analyzing: [ frac{a+1}{a(a+2)} + frac{b+1}{b(b+2)} + frac{c+1}{c(c+2)} geqslant frac{9}{ frac{a(a+2)}{a+1} + frac{b(b+2)}{b+1} + frac{c(c+2)}{c+1}} ] 2. We need to compute the harmonic sum and simplify it. Observing from part (1): [ frac{a(a+2)}{a+1} = a + 2 - frac{1}{a+1} ] 3. Extending the above simplification, we have: [ frac{a(a+2)}{a+1} + frac{b(b+2)}{b+1} + frac{c(c+2)}{c+1} = (a + 2 + b + 2 + c + 2) - left(frac{1}{a+1} + frac{1}{b+1} + frac{1}{c+1}right) ] 4. Extending this, [ frac{a(a+2)}{a+1} + frac{b(b+2)}{b+1} + frac{c(c+2)}{c+1} = (a + b + c + 6) - left(frac{1}{a+1} + frac{1}{b+1} + frac{1}{c+1}right) ] 5. Considering (a + b + c leqslant 3), it follows, [ a + b + c + 6 leqslant 9 ] 6. Therefore, [ frac{a(a+2)}{a+1} + frac{b(b+2)}{b+1} + frac{c(c+2)}{c+1} leqslant 9 - left( frac{1}{a+1} + frac{1}{b+1} + frac{1}{c+1} right) ] 7. Using frac{{1}{a+1} + frac{1}{b+1} + frac{1}{c+1} geqslant frac{3}{2}} from part (1), [ frac{a(a+2)}{a+1} + frac{b(b+2)}{b+1} + frac{c(c+2)}{c+1} leqslant 9 - frac{3}{2} = 7.5 ] 8. Thus, [ frac{9}{ frac{a(a+2)}{a+1} + frac{b(b+2)}{b+1} + frac{c(c+2)}{c+1}} geqslant frac{9}{7.5} = frac{6}{5} < 2 ] 9. Consequently, [ frac{a+1}{a(a+2)} + frac{b+1}{b(b+2)} + frac{c+1}{c(c+2)} geqslant 2 ]

answer:We need to find the derivative of the function: y = e^{ax} left( frac{1}{2a} + frac{a cos(2bx) + 2b sin(2bx)}{2(a^2 + 4b^2)} right) We denote: f(x) = e^{ax} g(x) = frac{1}{2a} + frac{a cos(2bx) + 2b sin(2bx)}{2(a^2 + 4b^2)} We will use the product rule, which states that: (f cdot g)' = f' cdot g + f cdot g' First, we compute the derivative of ( f(x) ): f'(x) = frac{d}{dx}e^{ax} = ae^{ax} Next, we need to find the derivative of ( g(x) ). We have: g(x) = frac{1}{2a} + frac{a cos(2bx) + 2b sin(2bx)}{2(a^2 + 4b^2)} For simplicity, let's write: g(x) = frac{1}{2a} + frac{h(x)}{2(a^2 + 4b^2)} where h(x) = a cos(2bx) + 2b sin(2bx) To find ( g'(x) ), we need the derivative of ( h(x) ). h'(x) = frac{d}{dx} (a cos(2bx) + 2b sin(2bx)) Using the chain rule: h'(x) = -a sin(2bx) cdot 2b + 2b cos(2bx) cdot 2b h'(x) = -2ab sin(2bx) + 4b^2 cos(2bx) Now, substituting back into ( g'(x) ): g'(x) = 0 + frac{h'(x)}{2(a^2 + 4b^2)} g'(x) = frac{-2ab sin(2bx) + 4b^2 cos(2bx)}{2(a^2 + 4b^2)} Simplifying, we get: g'(x) = frac{-ab sin(2bx) + 2b^2 cos(2bx)}{a^2 + 4b^2} Now, apply the product rule: y' = f'(x) cdot g(x) + f(x) cdot g'(x) y' = ae^{ax} left( frac{1}{2a} + frac{a cos(2bx) + 2b sin(2bx)}{2(a^2 + 4b^2)} right) + e^{ax} cdot frac{-ab sin(2bx) + 2b^2 cos(2bx)}{a^2 + 4b^2} First term: ae^{ax} left( frac{1}{2a} + frac{a cos(2bx) + 2b sin(2bx)}{2(a^2 + 4b^2)} right) = e^{ax} left( frac{a}{2a} + frac{a^2 cos(2bx) + 2ab sin(2bx)}{2(a^2 + 4b^2)} right) = e^{ax} left( frac{1}{2} + frac{a^2 cos(2bx) + 2ab sin(2bx)}{2(a^2 + 4b^2)} right) Second term: e^{ax} cdot frac{-ab sin(2bx) + 2b^2 cos(2bx)}{a^2 + 4b^2} Combining both: y' = e^{ax} left( frac{1}{2} + frac{a^2 cos(2bx) + 2ab sin(2bx)}{2(a^2 + 4b^2)} right) + e^{ax} cdot frac{-ab sin(2bx) + 2b^2 cos(2bx)}{a^2 + 4b^2} To simplify: y' = frac{e^{ax}}{2} + frac{e^{ax}}{2(a^2 + 4b^2)} (a^2 cos(2bx) + 2ab sin(2bx) - ab sin(2bx) + 2b^2 cos(2bx)) = frac{e^{ax}}{2} + frac{e^{ax}}{2(a^2 + 4b^2)} (a^2 cos(2bx) + 4b^2 cos(2bx)) = frac{e^{ax}}{2} + frac{e^{ax}}{2(a^2 + 4b^2)} (a^2 + 4b^2) cos(2bx) = frac{e^{ax}}{2} + frac{e^{ax}}{2} cos(2bx) = frac{e^{ax}}{2} (1 + cos(2bx)) = e^{ax} cdot left( frac{1 + cos(2bx)}{2} right) Using the double-angle identity ( cos(2u) = 2cos^2(u) - 1 ): frac{1 + cos(2bx)}{2} = cos^2(bx) Thus, we have: y' = e^{ax} cdot cos^2(bx) Conclusion: boxed{e^{ax} cos^2(bx)}