answer:To find the minimum value of the given expression over all real values x, we start by introducing a substitution to simplify the expression. Let t = cos^2 x. Consequently, we have sin^2 x = 1 - t. Substituting these into the original expression, we get: begin{align*} frac{sin^6 x + cos^6 x + 1}{sin^4 x + cos^4 x + 1} &= frac{(1 - t)^3 + t^3 + 1}{(1 - t)^2 + t^2 + 1} &= frac{1 - 3t + 3t^2 - t^3 + t^3 + 1}{1 - 2t + t^2 + t^2 + 1} &= frac{3t^2 - 3t + 2}{2t^2 - 2t + 2}. end{align*} Next, we simplify the fraction by dividing the numerator and the denominator by the coefficient of the quadratic term in the denominator, which leads to: [frac{3t^2 - 3t + 2}{2t^2 - 2t + 2} = frac{3}{2} - frac{1}{2(t^2 - t + 1)}.] To minimize this expression, we need to maximize the denominator of the subtracted fraction, which is equivalent to minimizing t^2 - t + 1. The minimum value of a quadratic function at^2 + bt + c occurs at t = -frac{b}{2a}. Applying this to our quadratic gives t = frac{1}{2}. Substituting t = frac{1}{2} into the expression, we find the minimum value: [frac{3}{2} - frac{1}{2left(left(frac{1}{2}right)^2 - frac{1}{2} + 1right)} = frac{3}{2} - frac{1}{2left(frac{1}{4} - frac{1}{2} + 1right)} = frac{3}{2} - frac{1}{2left(frac{3}{4}right)} = frac{3}{2} - frac{1}{frac{3}{2}} = frac{3}{2} - frac{2}{3} = frac{9 - 4}{6} = frac{5}{6}.] Therefore, the minimum value of the given expression over all real values x is boxed{frac{5}{6}}.

answer:Solution: (1) Proof: Given 3a_na_{n-1}+a_n-a_{n-1}=0 (ngeq2), we have a_{n-1}-a_n=3a_na_{n-1}, thus frac{a_{n-1}-a_n}{a_na_{n-1}}=frac{1}{a_n}-frac{1}{a_{n-1}}=3 (ngeq2). Therefore, the sequence left{frac{1}{a_n}right} is an arithmetic sequence with the first term 1 and common difference 3. (2) From (1), we have frac{1}{a_n}=1+3(n-1)=3n-2, thus a_n=boxed{frac{1}{3n-2}}.

answer:The single digit numbers between 4 and 9 are 5, 6, 7, and 8. Since the number is less than 6, the only number that fits this description is boxed{5} .

answer:To find the largest four-digit number whose digits add up to 16, we start by maximizing the leftmost digit, which is the thousandth place. This is because larger values in the leftmost positions contribute more to the overall size of the number. 1. Assign the maximum possible value to the thousandth digit without exceeding a total sum of 16 for all digits. This value is 9, as it is the largest single-digit number. 2. Subtract the value of the thousandth digit from the total desired sum: 16 - 9 = 7. This leaves us with 7 to be distributed among the remaining three digits. 3. To keep the number as large as possible, we assign the largest possible value to the next significant digit, the hundredth place. Since we have 7 left and we need to distribute it among three digits, the largest value we can assign to the hundredth place without exceeding our total is 7 itself. 4. This leaves 0 to be distributed among the last two digits, the tenth and the unit places. Therefore, both are assigned 0. Following these steps, the largest four-digit number whose digits add up to 16 is constructed as 9700. Thus, the final answer is encapsulated as boxed{9700}.