answer:1. We will solve the expression step by step, following the order of operations (PEMDAS/BODMAS). (1)(0.25)^{frac{1}{2}} - [-2 times (frac{3}{7})^0]^2 times [(-2)^3]^{frac{4}{3}} + (sqrt{2} - 1)^{-1} - 2^{frac{1}{2}} = (0.25)^{frac{1}{2}} - [-2 times 1]^2 times [(-2)^3]^{frac{4}{3}} + (sqrt{2} - 1)^{-1} - 2^{frac{1}{2}} = frac{1}{2} - (-2)^2 times [(-2)^3]^{frac{4}{3}} + (sqrt{2} - 1)^{-1} - sqrt{2} = frac{1}{2} - 4 times (8^{frac{4}{3}}) + (sqrt{2} - 1)^{-1} - sqrt{2} = frac{1}{2} - 4 times 16 + frac{1}{sqrt{2} - 1} - sqrt{2} = frac{1}{2} - 64 + frac{sqrt{2} + 1}{(sqrt{2} - 1)(sqrt{2} + 1)} - sqrt{2} = frac{1}{2} - 64 + sqrt{2} + 1 - sqrt{2} = -frac{125}{2} So, the value of the expression is boxed{-frac{125}{2}}. 2. We will solve the expression step by step, using logarithmic properties. (log_5 5)^2 + log_2 times log_50 = (1)^2 + log_2 2 times (log_5 10 + log_5 5) = 1 + 1 times (1 + log_5 5) = 1 + 1 + 1 = 3 So, the value of the expression is boxed{3}.

answer:Since alpha is the largest interior angle of the triangle, it follows that frac{pi}{3} < alpha < pi, and frac{2pi}{3} < 2alpha < 2pi, Since cos 2alpha = frac{1}{2} > 0, it follows that frac{3pi}{2} < 2alpha < 2pi, solving this gives: frac{3pi}{4} < alpha < pi, Therefore, tanalpha < 0, Since cos 2alpha = frac{cos^2alpha - sin^2alpha}{cos^2alpha + sin^2alpha} = frac{1 - tan^2alpha}{1 + tan^2alpha} = frac{1}{2}, solving this gives: tan^2alpha = frac{1}{3}, that is: tanalpha = -frac{sqrt{3}}{3}, Therefore, frac{1 - tanalpha}{1 + tanalpha} = frac{1 - (-frac{sqrt{3}}{3})}{1 + (-frac{sqrt{3}}{3})} = 2 + sqrt{3}. Hence, the correct choice is boxed{B}. From the problem, we can deduce that frac{pi}{3} < alpha < pi, which allows us to determine the range of 2alpha. Combining this with cos 2alpha = frac{1}{2} > 0, we can solve for the range frac{3pi}{4} < alpha < pi. Using the double-angle cosine formula and the basic trigonometric identity, we can solve for tan^2alpha = frac{1}{3}, which allows us to find the value of tanalpha, and thus calculate the value of frac{1 - tanalpha}{1 + tanalpha}. This problem mainly examines the application of the double-angle cosine formula, basic trigonometric identities, and the properties and graphs of cosine functions in simplifying and evaluating trigonometric expressions. It tests the concept of transformation and the integration of mathematical and geometric thinking, and is considered a medium-level question.

answer:1. **Simplify the given expression**: Start by simplifying each part of the expression: [ left(frac{(x+1)^3(x^2-x+1)^3}{(x^3+1)^3}right)^3 cdot left(frac{(x-1)^3(x^2+x+1)^3}{(x^3-1)^3}right)^3 ] We can simplify each fraction: [ left(frac{(x+1)(x^2-x+1)}{x^3+1}right)^9 cdot left(frac{(x-1)(x^2+x+1)}{x^3-1}right)^9 ] 2. **Factorize and simplify further**: Since x^3 + 1 can be factorized as (x+1)(x^2-x+1) and x^3 - 1 can be factorized as (x-1)(x^2+x+1): [ left(frac{x^3+1}{x^3+1}right)^9 cdot left(frac{x^3-1}{x^3-1}right)^9 ] 3. **Evaluate the simplified expression**: Since frac{x^3+1}{x^3+1} = 1 and frac{x^3-1}{x^3-1} = 1 everywhere except possibly at x neq -1 or x neq 1: [ 1^9 cdot 1^9 = 1 ] 4. **Conclude**: The expression simplifies to 1 for all x except possibly x = -1 or x = 1, but the limits at these points still result in the value 1. Thus, the answer is 1. The final answer is boxed{C}

answer:There are 4 different cases for the starting lineup: **Case 1: Bob starts but not Yogi and Zane.** The coach must choose 4 more players from the 12 remaining players (excluding Yogi and Zane). Thus, there are: binom{12}{4} **Case 2: Yogi starts but not Bob and Zane.** Similar to Case 1, the coach must choose 4 more players from the 12 remaining players (excluding Bob and Zane). Hence: binom{12}{4} **Case 3: Zane starts but not Bob and Yogi.** As in previous cases, the coach chooses 4 more players from the 12 remaining. Therefore: binom{12}{4} **Case 4: Neither Bob, Yogi, nor Zane starts.** Here, the coach chooses all 5 players from the 12 remaining players. So: binom{12}{5} Calculating these, we have: [ binom{12}{4} = frac{12 times 11 times 10 times 9}{4 times 3 times 2 times 1} = 495 ] [ binom{12}{5} = frac{12 times 11 times 10 times 9 times 8}{5 times 4 times 3 times 2 times 1} = 792 ] Adding these up gives: [ 495 + 495 + 495 + 792 = 2277 ] So, the total number of possible starting lineups is boxed{2277}.