answer:Obviously, 1 is not a "wise number", and odd numbers greater than 1, 2k+1=(k+1)^{2}-k^{2}, are all "wise numbers".From 4k=(k+1)^{2}-(k-1)^{2}, we know that numbers greater than 4 and divisible by 4 are all "wise numbers", but 4 is not a "wise number".Since x^{2}-y^{2}=(x+y)(x-y)(x, y in mathbf{N}), when x and y have the same parity, (x+y)(x-y) is divisible by 4. When x and y have different parity, (x+y)(x-y) is odd. Therefore, numbers of the form 4k+2 are not "wise numbers".In the sequence of natural numbers, among the first four natural numbers, only 3 is a "wise number". Thereafter, in every consecutive four integers, there are three "wise numbers".Since 1989=3 times 663, 2656=4 times 664 is the 1990th "wise number".

answer:Proof: By induction, it is easy to see that a_{0}frac{1}{a_{n-1}}-frac{1}{n}>frac{1}{a_{n-2}}-frac{2}{n}>cdots>frac{1}{a_{0}}-1=1, i.e., a_{n}a_{k}left(1+frac{a_{k+1}}{n+1}right). Therefore, we havefrac{1}{a_{k+1}}frac{n+1}{n+2}=1-frac{1}{n+2}>1-frac{1}{n}.In summary, 1-frac{1}{n}<a_{n}<1.

answer:SOLUTION. Neither x nor y can be zero, as they are in the denominators of the fractions in the given inequality. The problem does not specify the sign of the numbers x and y, so we must not forget that these numbers can also be negative when performing transformations.We will first try to simplify the expressions in the problem. By multiplying the inequality by the positive number x^{2} y^{2}, we can equivalently eliminate the denominators:begin{aligned}& (x+y) frac{x+y}{x y} geqqleft(frac{x^{2}+y^{2}}{x y}right)^{2}, quad mid cdot x^{2} y^{2} & x y(x+y)^{2} geqqleft(x^{2}+y^{2}right)^{2} .end{aligned}Now it is clear that by expanding the expressions in the obtained inequality, we will get the term 2 x^{2} y^{2} on both sides - we can cancel this term in the first step and further transform the inequality into a product form:begin{aligned}x^{3} y+2 x^{2} y^{2}+x y^{3} & geqq x^{4}+2 x^{2} y^{2}+y^{4}, x^{3} y+x y^{3} & geqq x^{4}+y^{4}, 0 & geqq x^{4}-x^{3} y+y^{4}-x y^{3}, 0 & geqq x^{3}(x-y)-y^{3}(x-y), 0 & geqqleft(x^{3}-y^{3}right)(x-y) .end{aligned}On the right side of the last inequality, we have the product of two binomials, and since we are comparing it to zero, it is sufficient to examine the signs of the individual parentheses.If x geqq y, then x^{3} geqq y^{3}, and similarly, if x leqq y, then x^{3} leqq y^{3} (this is a consequence of the fact that the function of the third power is increasing over the entire domain of real numbers). Therefore, the expressions in the parentheses have the same sign for any values of x and y, so the inequality left(x^{3}-y^{3}right)(x-y) geqq 0 holds. Inequality (1) can only be satisfied if one of the parentheses is zero. The equations x=y and x^{3}=y^{3} are equivalent, so inequality (1) holds if and only if x=y.The solution consists of all pairs of real numbers (x, y) where x=y neq 0. In the transformations, we used only equivalent transformations, so a verification of correctness is not necessary.Note. The product on the right side of (1) can also be transformed using the formula x^{3}-y^{3}=(x-y)left(x^{2}+x y+y^{2}right) into the product of two non-negative polynomials:left(x^{3}-y^{3}right)(x-y)=(x-y)^{2}left(x^{2}+x y+y^{2}right) .The trinomial in the second parenthesis has a non-positive discriminant D(y)=y^{2}-4 y^{2}=-3 y^{2} leqq 0 for any y. Therefore, the equality x^{2}+x y+y^{2}=0 occurs only in the case x=y=0.## GUIDING AND SUPPLEMENTARY PROBLEMS:N1. Determine all pairs (x, y) of positive real numbers that satisfy the inequality x / y + y / x leqq 2. [After eliminating the fractions, we transform the inequality to (x-y)^{2} leqq 0, which holds only for x=y.]N2. Determine all pairs (x, y) of real numbers that satisfy the inequality x / y + y / x leqq 2. [The solution consists of all pairs (x, y) such that x=y neq 0 or x y<0.]N3. Determine all pairs (x, y) of real numbers that satisfy the inequality x^{2}+4 y^{2} leqq 4 x y. [The solution consists of all pairs (x, y) such that x=2 y.]N4. Prove that for any real numbers x, y, the inequality x^{4}+y^{4} geqq x^{3} y+x y^{3} holds. [Transformation into a product.]D1. Prove that for any positive numbers a, b, the inequalitysqrt[3]{frac{a}{b}}+sqrt[3]{frac{b}{a}} leqq sqrt[3]{2(a+b)left(frac{1}{a}+frac{1}{b}right)}holds. Determine when equality occurs. [49-A-II-3]

answer:1. If f(1) = f(2) = f(3) = 1 or 2 or 3, there are a total of 3 functions.2. If f(1) = 1; f(2) = f(3) = 2 or 3, there are a total of 2 functions. If f(2) = 2; f(1) = f(3) = 1 or 3, there are a total of 2 functions. If f(3) = 3; f(1) = f(2) = 1 or 2, there are a total of 2 functions. 3. If f(1) = 1; f(2) = 2; f(3) = 3, there is 1 function. Therefore, there are a total of 10 such functions. Hence, the correct option is boxed{text{D}}.

answer:Given: 204 = 2 times 85 + 34 85 = 2 times 34 + 17 34 = 2 times 17 Therefore, the greatest common divisor of 204 and 85 is 17. Hence, the answer is: boxed{17}. By dividing the larger number by the smaller one, obtaining a quotient and a remainder, and then dividing the divisor from the previous step by the new remainder (choosing the larger number to divide the smaller one each time), this process is repeated until a division is exact. This method is known as the Euclidean algorithm for finding the greatest common divisor of two numbers. This problem tests the application of the Euclidean algorithm, and it is important to perform the calculations accurately as it is a basic question.

answer:5. Solution: Let Aleft(x_{1}, y_{1}right), Bleft(-k y_{1}, k x_{1}right), then we have |O A|^{2}=x_{1}^{2}+y_{1}^{2}, |O B|^{2}=k^{2}|O A|^{2}, and from frac{x_{1}^{2}}{a^{2}}+frac{y_{1}^{2}}{b^{2}}=1, frac{y_{1}^{2}}{a^{2}}+frac{x_{1}^{2}}{b^{2}}=frac{1}{k^{2}} adding them together we get: left(frac{1}{a^{2}}+frac{1}{b^{2}}right) |O A|^{2}=frac{k^{2}+1}{k^{2}}, |O A|^{2}=frac{a^{2} b^{2}}{a^{2}+b^{2}} frac{k^{2}+1}{k^{2}}, the distance from point O to line A B is d^{2}=frac{|O A|^{2}|O B|^{2}}{|A B|^{2}}=frac{|O A|^{2}|O B|^{2}}{|O A|^{2}+|O B|^{2}}= frac{k^{2}|O A|^{2}}{1+k^{2}}=frac{a^{2} b^{2}}{a^{2}+b^{2}}, d=frac{a b}{sqrt{a^{2}+b^{2}}}.