answer:To solve the given problem, we start by applying the operation defined for a second-order determinant to the determinant provided: [|begin{array}{l}{x+3}&{x-3}{x-3}&{x+3}end{array}| = (x+3)(x+3) - (x-3)(x-3)] Expanding both squares and simplifying, we get: [(x+3)^2 - (x-3)^2 = x^2 + 6x + 9 - (x^2 - 6x + 9) = 12x] According to the problem, this expression equals 12: [12x = 12] To find x, we divide both sides by 12: [x = frac{12}{12}] Simplifying the fraction gives us: [x = 1] Therefore, the value of x that satisfies the given determinant equation is boxed{1}.

answer:Solution: (1) The function f(x) is monotonically increasing on the domain (0,+infty). Proof: Let 0 < x_{1} < x_{2}, then dfrac{x_{2}}{x_{1}} > 1, Since when x > 1, f(x) > 0 always holds, and f(x)+f(dfrac{1}{x})=0, Therefore, f(x_{2})-f(x_{1})=f(x_{2})+f(dfrac{1}{x_{1}})=f(dfrac{x_{2}}{x_{1}}) > 0, Thus, f(x_{1}) < f(x_{2}), Therefore, the function f(x) is monotonically increasing on the domain (0,+infty); (2) Since f(x)+f(x-2) leqslant 3=f(8), and the function f(x) is monotonically increasing on the domain (0,+infty), Therefore, begin{cases} x > 0 x-2 > 0 x(x-2)leqslant 8end{cases}, solving this yields: 2 < xleqslant 4, Therefore, the solution set of the inequality f(x)+f(x-2) leqslant 3 is boxed{{x|2 < xleqslant 4}}.

answer:1. **Identify the Triangle Type**: Given the side lengths 5, 5, 6, this is an isosceles triangle where AB = AC = 5 and BC = 6. 2. **Determine Midpoints and Crease Formation**: The midpoint, M, of BC divides it into two segments each of length 3 inches. When vertex A is folded onto point M, crease AM is formed. 3. **Calculate Length of AM Using the Triangle**: In triangle ABM, where AB = 5, BM = 3, and AM is the crease to be calculated. - **Apply the Law of Cosines in triangle ABM**: [ AM^2 = AB^2 + BM^2 - 2 cdot AB cdot BM cdot cos(angle ABM) ] Since angle ABM = 90^circ (as AB is a radius and BM half the chord length, forming a right triangle), [ AM^2 = 5^2 + 3^2 - 2 cdot 5 cdot 3 cdot 0 ] [ AM^2 = 25 + 9 = 34 ] [ AM = sqrt{34} ] 4. **Conclusion with Boxed Answer**: [ sqrt{34} ] The final answer is boxed{sqrt{34}} (Choice C).

answer:Given a complete graph ( K_{6} ) with edges colored using two colors, we are to prove that there exists at least one monochromatic triangle, i.e., a triangle where all three edges are the same color. 1. **Setup:** - Let the 6 vertices be ( A_{i} ) where ( i = 1, 2, ldots, 6 ). - Consider any vertex, say ( A_{1} ). 2. **Initial Edge Consideration:** - From ( A_{1} ), there are 5 edges connecting to the other 5 vertices. Since these edges can only be colored with two colors, by the pigeonhole principle (or Dirichlet's principle), at least 3 of these 5 edges must be of the same color. Let's assume without loss of generality that these 3 edges are red and they connect ( A_{1} ) to vertices ( A_{2}, A_{3}, ) and ( A_{4} ). 3. **Subgraph Analysis:** - Now, consider the subgraph formed by vertices ( A_{2}, A_{3}, ) and ( A_{4} ). There are three edges within this subgraph: ( A_{2}A_{3}, A_{2}A_{4}, ) and ( A_{3}A_{4} ). Since these three vertices all connect to ( A_{1} ) via red edges, we need to analyze the colors of the edges between ( A_{2}, A_{3}, ) and ( A_{4} ). 4. **Monochromatic Triangle Formation:** - If any of the edges ( A_{2}A_{3}, A_{2}A_{4}, ) or ( A_{3}A_{4} ) is red, assume without loss of generality that ( A_{2}A_{3} ) is red, then ( triangle A_{1}A_{2}A_{3} ) forms a monochromatic red triangle. - If none of the edges among ( A_{2}A_{3}, A_{2}A_{4}, ) or ( A_{3}A_{4} ) is red, then all of these three edges must be blue. Hence, ( triangle A_{2}A_{3}A_{4} ) forms a monochromatic blue triangle. 5. **Conclusion:** - Hence, every scenario concludes with at least one monochromatic triangle in the complete graph ( K_{6} ). Thus, it is proven that any 2-coloring of the edges of ( K_6 ) must result in at least one monochromatic triangle. (blacksquare)