answer:1. **Rationalize each term**: Rationalize by multiplying numerator and the denominator by the conjugate of the denominator: [frac{1}{4-sqrt{15}} = frac{4+sqrt{15}}{(4-sqrt{15})(4+sqrt{15})} = frac{4+sqrt{15}}{16-15} = 4 + sqrt{15}.] [frac{1}{sqrt{15} - 2sqrt{3}} = frac{sqrt{15} + 2sqrt{3}}{15 - 12} = sqrt{15} + 2sqrt{3}.] [frac{2}{2sqrt{3} - sqrt{12}} = frac{2(2sqrt{3} + sqrt{12})}{12 - 12} = 2(2sqrt{3} + 2sqrt{3}) = 8sqrt{3}.] [frac{1}{sqrt{12} - 3} = frac{sqrt{12} + 3}{12 - 9} = sqrt{12} + 3.] [frac{1}{3 - sqrt{8}} = frac{3 + sqrt{8}}{9 - 8} = 3 + sqrt{8}.] 2. **Combine all terms**: [ T = (4 + sqrt{15}) - (sqrt{15} + 2sqrt{3}) + 8sqrt{3} - (sqrt{12} + 3) + (3 + sqrt{8}). ] Now, group and simplify terms: [ T = 4 - 2sqrt{3} + 8sqrt{3} - sqrt{12} + sqrt{8} + 3 = 7 + 6sqrt{3} - 2sqrt{3} + sqrt{8}. ] Simplifying: [ T = 7 + 4sqrt{3} + 2sqrt{2}. ] 3. **Conclusion**: [ T = 7 + 4sqrt{3 + 2sqrt{2}} ] The final answer is **A.** boxed{7 + 4sqrt{3} + 2sqrt{2}}

answer:1. Let ( |A cap B cap C| = x ). We need to find the possible values of ( x ). 2. Using the principle of inclusion-exclusion for three sets, we have: [ |A cup B cup C| = |A| + |B| + |C| - |A cap B| - |A cap C| - |B cap C| + |A cap B cap C| ] Substituting the given values: [ |A cup B cup C| = 92 + 35 + 63 - 16 - 51 - 19 + x ] Simplifying the expression: [ |A cup B cup C| = 104 + x ] 3. Since ( |A cup B cup C| ) must be a non-negative integer, we have: [ 104 + x geq 0 implies x geq -104 ] However, this constraint is not useful since ( x ) must be a non-negative integer. 4. Next, consider the individual intersections: - For ( |A cap B| = 16 ): [ |A cap B| = |(A cap B) setminus C| + |A cap B cap C| = 16 - x ] Since ( |(A cap B) setminus C| geq 0 ): [ 16 - x geq 0 implies x leq 16 ] - For ( |A cap C| = 51 ): [ |A cap C| = |(A cap C) setminus B| + |A cap B cap C| = 51 - x ] Since ( |(A cap C) setminus B| geq 0 ): [ 51 - x geq 0 implies x leq 51 ] - For ( |B cap C| = 19 ): [ |B cap C| = |(B cap C) setminus A| + |A cap B cap C| = 19 - x ] Since ( |(B cap C) setminus A| geq 0 ): [ 19 - x geq 0 implies x leq 19 ] 5. Combining these constraints, we have: [ x leq min(16, 51, 19) = 16 ] 6. Now, consider the constraint from the total number of elements in ( A ): [ |A| = |A setminus (B cup C)| + |A cap B| + |A cap C| - |A cap B cap C| ] Substituting the given values: [ 92 = |A setminus (B cup C)| + 16 + 51 - x ] Simplifying: [ 92 = |A setminus (B cup C)| + 67 - x ] [ |A setminus (B cup C)| = 25 + x ] Since ( |A setminus (B cup C)| geq 0 ): [ 25 + x geq 0 implies x geq -25 ] This constraint is not useful since ( x ) must be a non-negative integer. 7. Similarly, consider the constraint from the total number of elements in ( B ): [ |B| = |B setminus (A cup C)| + |A cap B| + |B cap C| - |A cap B cap C| ] Substituting the given values: [ 35 = |B setminus (A cup C)| + 16 + 19 - x ] Simplifying: [ 35 = |B setminus (A cup C)| + 35 - x ] [ |B setminus (A cup C)| = x ] Since ( |B setminus (A cup C)| geq 0 ): [ x geq 0 ] 8. Finally, consider the constraint from the total number of elements in ( C ): [ |C| = |C setminus (A cup B)| + |A cap C| + |B cap C| - |A cap B cap C| ] Substituting the given values: [ 63 = |C setminus (A cup B)| + 51 + 19 - x ] Simplifying: [ 63 = |C setminus (A cup B)| + 70 - x ] [ |C setminus (A cup B)| = x - 7 ] Since ( |C setminus (A cup B)| geq 0 ): [ x - 7 geq 0 implies x geq 7 ] 9. Combining all the constraints, we have: [ 7 leq x leq 16 ] 10. Therefore, the number of possible values for ( x ) is: [ 16 - 7 + 1 = 10 ] The final answer is (boxed{10})

answer:To find the slope of the tangent line at the point (1, f(1)), we need to calculate the derivative of f(x) with respect to x. Let's find the derivative of f(x): f'(x) = frac{d}{dx}left(2ax^2 - frac{1}{ax}right) = frac{d}{dx}(2ax^2) - frac{d}{dx}left(frac{1}{ax}right) = 4ax - frac{1}{ax^2}. Now, calculate the slope at the point where x=1: k = f'(1) = 4a(1) - frac{1}{a(1)^2} = 4a - frac{1}{a}. Since a > 0, we can apply the AM-GM inequality, which states that the arithmetic mean is greater than or equal to the geometric mean: k = 4a + frac{1}{a} geq 2sqrt{4a cdot frac{1}{a}} = 2sqrt{4} = 4. Equality holds when 4a = frac{1}{a}, which gives a^2 = frac{1}{4}, so a = frac{1}{2} since a > 0. Therefore, the minimum value of k is 4 when a = frac{1}{2}. The final answer is that the minimum value of k is attained when a = boxed{frac{1}{2}}.

answer:Since sin theta = 1 - log_{2}x in [-1,1], we have 0 leqslant log_{2}x leqslant 2. Solving this, we get 1 leqslant x leqslant 4. Therefore, the answer is: boxed{[1,4]}. Based on sin theta = 1 - log_{2}x in [-1,1], we can derive that 0 leqslant log_{2}x leqslant 2, from which we can find the range of x. This question mainly examines the solution of logarithmic inequalities, and it is a basic problem.