answer:To solve the given problem, we analyze the function f(x) based on the given functional equation f(xy) = y^2f(x) + x^2f(y), with the domain of f(x) being mathbb{R}. We will examine each statement by substituting specific values into the equation. **For Statement A: f(0)=0** Let's substitute x=0 and y=0 into the functional equation: begin{align*} f(0cdot0) &= 0^2f(0) + 0^2f(0) f(0) &= 0 + 0 f(0) &= 0 end{align*} This confirms that statement A is true. **For Statement B: f(1)=0** Now, let's substitute x=1 and y=1: begin{align*} f(1cdot1) &= 1^2f(1) + 1^2f(1) f(1) &= 2f(1) end{align*} Subtracting f(1) from both sides gives 0 = f(1), which means statement B is also true. **For Statement C: f(x) is an even function** To check if f(x) is even, we need to show that f(-x) = f(x) for all x. Let's substitute y=-1: begin{align*} f(-x) &= (-1)^2f(x) + x^2f(-1) f(-x) &= f(x) + x^2f(-1) end{align*} From the previous steps, we found that f(1) = 0 and substituting x=y=-1 gives f(-1) = frac{1}{2}f(1) = 0. Thus, f(-x) = f(x), confirming that f(x) is an even function. Hence, statement C is true. **For Statement D: x=0 is a local minimum point of f(x)** Given that f(0)=0 and assuming f(x)=0 for all x (which satisfies the given functional equation), the function f(x)=0 is constant and does not have any extremum points. Therefore, statement D cannot be confirmed as true without additional information about the behavior of f(x) around x=0 beyond it being constant in our assumption. In conclusion, the correct choices based on the given functional equation and the analysis are: boxed{ABC}

answer:To solve this problem, let's consider two distinct scenarios based on the conditions provided: 1. **Classmates A and B are both invited:** We then need to choose 4 additional classmates from the remaining 8 (excluding A and B) to join the event. The number of ways to select 4 out of 8 is given by the combination formula C(8, 4). This yields: C(8, 4) = frac{8!}{4!(8-4)!} = frac{8 times 7 times 6 times 5}{4 times 3 times 2 times 1} = 70 2. **Neither classmate A nor B is invited:** Now we need to pick 6 classmates from the remaining 8 to invite. The number of ways to do this is given by C(8, 6), which is the same as C(8, 2) due to the symmetric property of combinations: C(8, 6) = C(8, 2) = frac{8!}{2!(8-2)!} = frac{8 times 7}{2 times 1} = 28 Adding the counts from both scenarios gives us the total number of different ways to extend the invitation: 70 + 28 = 98 Therefore, the student has 98 different methods to invite their classmates under the given constraints. boxed{98} methods of invitation can be used.

answer:First, analyze the given function. Based on its analytic expression, determine the parity and monotonicity of the function. Then, solve the inequality using the properties of even functions. The function is defined as fleft(xright)= ln left(1+left|xright|right)-frac{1}{1+{x}^{2}}. By observing the function, we can see that fleft(-xright)=fleft(xright), which means the function is even and symmetric about the y-axis. When x > 0, the function is monotonically increasing. Now, to find the range of x that satisfies fleft( x right) > fleft( 2x-1 right), we can start by finding the range of x where left|xright| > left|2x-1right|. Squaring both sides and solving the inequality, we get frac{1}{3} < x < 1. Therefore, the correct answer is boxed{A:  left( frac{1}{3},1 right)}.

answer:First, simplify the expression inside the sum: [ sqrt{n + sqrt{n^2 - 4}} = frac{1}{sqrt{2}}sqrt{2n + 2sqrt{n^2 - 4}} = frac{1}{sqrt{2}}left(sqrt{n+2}+sqrt{n-2}right) ] Thus, the sum becomes: [ sum_{n = 1}^{10000} frac{1}{sqrt{n + sqrt{n^2 - 4}}} = sqrt{2}sum_{n = 1}^{10000} frac{1}{sqrt{n+2}+sqrt{n-2}} ] This is also a telescoping series: [ = frac{1}{sqrt{2}}sum_{n = 1}^{10000} left(sqrt{n+2}-sqrt{n-2}right) ] Evaluating the telescoping series, the sum simplifies to: [ frac{1}{sqrt{2}}left(sqrt{10002}+sqrt{10000}-sqrt{3}-sqrt{1}right) ] Calculating the remaining terms: [ = frac{1}{sqrt{2}}left(100 + 99.01 - 1.732 - 1right) approx frac{1}{sqrt{2}} times 196.278 = 70 + 88sqrt{2} ] Thus, ( p = 70 ), ( q = 88 ), and ( r = 2 ), and ( p+q+r = boxed{160} ).