answer:Using the double angle formulas, (sin x = 2 sin frac{x}{2} cos frac{x}{2}) and (cos x = 1 - 2 sin^2 frac{x}{2}), we can rewrite the expression: [ frac{1 + 2 sin frac{x}{2} cos frac{x}{2} + 1 - 2 sin^2 frac{x}{2}}{1 - 2 sin frac{x}{2} cos frac{x}{2} + 1 - 2 sin^2 frac{x}{2}} ] Simplify the expression: [ = frac{2 + 2 sin frac{x}{2} cos frac{x}{2} - 2 sin^2 frac{x}{2}}{2 - 2 sin frac{x}{2} cos frac{x}{2} - 2 sin^2 frac{x}{2}} ] Factor out the common terms: [ = frac{2(1 + sin frac{x}{2} cos frac{x}{2} - sin^2 frac{x}{2})}{2(1 - sin frac{x}{2} cos frac{x}{2} - sin^2 frac{x}{2})} ] [ = frac{1 + sin frac{x}{2} cos frac{x}{2} - sin^2 frac{x}{2}}{1 - sin frac{x}{2} cos frac{x}{2} - sin^2 frac{x}{2}} ] Using the identity (cos frac{x}{2} - sin frac{x}{2} = cos frac{x}{2} + sin frac{x}{2}): [ = frac{cos frac{x}{2}}{sin frac{x}{2}} ] [ = boxed{cot frac{x}{2}}. ]

answer:Let n=1, we get S_1 = 3x + 1; let n=2, we get S_2 = 9x + 1; let n=3, we get S_3 = 27x + 1, So a_1 = S_1 = 3x + 1, a_2 = S_2 - S_1 = 6x, a_3 = S_3 - S_2 = 18x Since {a_n} is a geometric sequence, we have a_2^2 = a_1 cdot a_3, Then (6x)^2 = 18x(3x + 1) Solving this, we get 18x(x + 1) = 0, Thus, x = 0 (discard) or x = -1, Therefore, x = -1 Hence, the answer is boxed{-1}.

answer:# Solution: Part 1: Mode of the Seventh-Grade Contestants' Scores The mode is the value that appears most frequently in a data set. For the seventh-grade contestants, we have the scores: 3, 6, 7, 6, 6, 8, 6, 9, 6, 10. Observing these scores, we see that the score 6 appears 5 times, which is more than any other score. Therefore, the mode of the seventh-grade contestants' scores is 6. boxed{6} Part 2: Average Score of the Eighth-Grade Contestants To find the average score, we sum all the scores and divide by the number of scores. For the eighth-grade contestants, the scores are: 5, 6, 8, 7, 5, 8, 7, 9, 8, 8. The sum of these scores is 2 times 5 + 6 + 2 times 7 + 4 times 8 + 9 = 10 + 6 + 14 + 32 + 9 = 71. There are 10 contestants, so the average score is: frac{1}{10} times 71 = 7.1 boxed{7.1} Part 3: Performance Analysis of Xiao Li and Xiao Zhang - **Median of Seventh-Grade Contestants' Scores**: To find the median, we arrange the scores in ascending order: 3, 6, 6, 6, 6, 7, 8, 9, 10. With 10 scores, the median is the average of the 5^{th} and 6^{th} scores, which are both 6. So, the median is frac{6+6}{2} = 6. - **Median of Eighth-Grade Contestants' Scores**: Arranging the eighth-grade scores in ascending order gives us: 5, 5, 6, 7, 7, 8, 8, 8, 8, 9. The median is the average of the 5^{th} and 6^{th} scores, which are 7 and 8. Thus, the median is frac{7+8}{2} = 7.5. - **Analysis**: Xiao Li's score (7) is higher than the median of the seventh-grade contestants (6), so his performance is considered "above average." Xiao Zhang's score (7) is lower than the median of the eighth-grade contestants (7.5), so his performance is considered "below average." boxed{text{Xiao Li: Above Average, Xiao Zhang: Below Average}}

answer:To find the minimum value of ab given a and b are positive and ab=a+b+3, we proceed as follows: 1. Starting with the given equation, ab = a + b + 3. 2. Since a and b are positive, we can apply the AM-GM inequality: a+b geqslant 2sqrt{ab}. 3. Substituting a + b from the given equation into the AM-GM inequality, we obtain ab - 3 = a + b geqslant 2sqrt{ab}. 4. Rearranging the inequality gives us ab - 3 geqslant 2sqrt{ab}, which simplifies to ab geqslant 2sqrt{ab} + 3. 5. We can rewrite this as a quadratic inequality in terms of sqrt{ab}: (sqrt{ab} - 3)(sqrt{ab} + 1) geqslant 0. 6. This inequality shows that sqrt{ab} geqslant 3, since sqrt{ab} + 1 is always positive for positive a and b. 7. Squaring both sides to eliminate the square root gives ab geqslant 9. 8. It's noted that equality holds (meaning the minimum value of ab is actually reached) when a = b = 3, satisfying both the original equation and the derived inequality. Therefore, the minimum value of ab is boxed{9}.