answer:To find the coefficient of the x^2 term in the expansion of (1+x)^8(1-x), we look at the terms that contribute to x^2:- The term x^2 comes from the product of the x^2 term in the expansion of (1+x)^8 and the constant term in (1-x), which is binom{8}{2}x^2 cdot 1 = 28x^2.- The term x^2 also comes from the product of the x term in the expansion of (1+x)^8 and the -x term in (1-x), which is binom{8}{1}x cdot (-x) = -8x^2.Adding these contributions together, we get 28x^2 - 8x^2 = 20x^2.Therefore, the coefficient of the x^2 term is 20.Hence, the answer is: boxed{20}.The problem involves the application of the binomial theorem in a flexible manner, making it a foundational question.

answer:8. B.From the given, we haven(n+2 m+1)=2^{2016} times 5^{2015} text {. }Thus, one of n and n+2 m+1 is odd and the other is even.Therefore, one of n and n+2 m+1 is even, with the possibilities being 2^{2016}, 2^{2016} times 5, 2^{2016} times 5^{2}, cdots, 2^{2016} times 5^{2015}, for a total of 2016 cases. Swapping the order also yields 2016 cases. Therefore, set A has a total of 4032 elements.

answer:[Solution] In the original identity, let y=1, then we get the identitythat is squaref(x) equiv f(x) f(1)-f(x+1)+1, x in Q .f(x+1) equiv f(x)+1Thus, for all x in Q, n in Z, we have f(x+n)=f(x)+n.Therefore squaref(n)=f(1)+n-1=n+1Next, in the original identity, take x=frac{1}{n}, y=n, x in Z, we getfleft(frac{1}{n} cdot nright)=fleft(frac{1}{n}right) cdot f(n)-fleft(frac{1}{n}+nright)+1Thus 2=fleft(frac{1}{n}right)(n+1)-fleft(frac{1}{n}right)-n+1,which means fleft(frac{1}{n}right)=1+frac{1}{n}.Finally, take x=p, y=frac{1}{q}, p in Z, q in N, we getfleft(p cdot frac{1}{q}right)=f(p) cdot fleft(frac{1}{q}right)-fleft(p+frac{1}{q}right)+1 .This leads to fleft(frac{p}{q}right)=(p+1)left(frac{1}{q}+1right)-frac{1}{q}-p=frac{p}{q}+1Therefore, the only function f(x)=x+1 satisfies all the conditions in the problem.

answer:AnalysisThis question tests the ability to compare values using the properties of exponential and logarithmic functions. By understanding the value ranges of these functions, we can easily determine the result. This is a basic question.SolutionFrom the properties of the exponential function, we know (a = 3^{0.4} > 1).From the properties of the logarithmic function, we know (b = log_{4}0.3 c > b).Hence, the correct answer is boxed{text{D}}.

answer:## Solution.Let's introduce the substitution a^{x}=y. Then we have fleft(y^{2}+2 y-7right)y-5which isy^{2}+y-2>0Solving this, we get y inlangle-infty,-2rangle cuplangle 1, inftyrangle, and due to the condition left(^{*}right), we have y inlangle 1, inftyrangle.Therefore, the original inequality is equivalent to the inequality a^{x}>1.From the condition f(0) cdot f(1)1=a^{0}, it follows that x<0.The final solution to the given inequality is x inlangle-infty, 0rangle.## NATIONAL MATHEMATICS COMPETITION## 4th grade - high school - B variant## April 25, 2023.

answer:7^{1}=7,7^{2}=49,7^{3}=343,7^{4}=2401; the unit digits repeat in the pattern 7,9,3,1, ldots7^{14}=left(7^{4}right)^{3} times 7^{2} therefore d=9