answer:We start by examining each polynomial separately. For the polynomial 3x^3 - 6x^2 - 4x + 24 = 0: By Vieta's formulas, the sum of the roots, s, of a cubic equation ax^3 + bx^2 + cx + d = 0 is given by s = -frac{b}{a}. Therefore, the sum of the roots for this cubic equation is: [ s = -frac{-6}{3} = 2 ] For the polynomial 4x^3 + 8x^2 - 20x - 60 = 0: Similarly, by Vieta's formulas, the sum of the roots for this equation is: [ s = -frac{8}{4} = -2 ] Since the roots of the given equation consist of the roots of both these equations (because a product of two terms equals zero if and only if at least one of the terms equals zero), the sum of all roots is: [ 2 + (-2) = 0 ] The sum of all roots of the equation [(3x^3 - 6x^2 - 4x + 24)(4x^3 + 8x^2 - 20x - 60) = 0] is boxed{0}.

answer:To calculate the value of 8cos ^{2}25^{circ}-tan 40^{circ}-4, we proceed as follows: First, we use the double angle formula for cosine: [8cos ^{2}25^{circ}-tan 40^{circ}-4 = 8timesfrac{1+cos 50^{circ}}{2}-tan 40^{circ}-4.] Simplifying, we get: [= 4cos 50^{circ}-tan 40^{circ}.] Since cos 50^{circ} = sin 40^{circ} (because 50^{circ} + 40^{circ} = 90^{circ}), we can rewrite the expression as: [= 4sin 40^{circ}-tan 40^{circ}.] Expressing tan 40^{circ} as frac{sin 40^{circ}}{cos 40^{circ}}, we have: [= 4sin 40^{circ}-frac{sin 40^{circ}}{cos 40^{circ}}.] Combining the terms under a common denominator, we get: [= frac{4sin 40^{circ}cos 40^{circ}-sin 40^{circ}}{cos 40^{circ}}.] Using the double angle formula for sine, sin 2theta = 2sin theta cos theta, we can rewrite the numerator as 2sin 80^{circ}-sin 40^{circ}: [= frac{2sin 80^{circ}-sin 40^{circ}}{cos 40^{circ}}.] Since sin 80^{circ} = cos 10^{circ}, we have: [= frac{2cos 10^{circ}-sin 40^{circ}}{cos 40^{circ}}.] Expanding sin 40^{circ} using the angle sum formula, sin(30^{circ}+10^{circ}) = sin 30^{circ}cos 10^{circ} + cos 30^{circ}sin 10^{circ}, we get: [= frac{2cos 10^{circ}-(frac{1}{2}cos 10^{circ}+frac{sqrt{3}}{2}sin 10^{circ})}{cos 40^{circ}}.] Simplifying, we find: [= frac{frac{3}{2}cos 10^{circ}-frac{sqrt{3}}{2}sin 10^{circ}}{cos 40^{circ}}.] Multiplying the numerator and the denominator by sqrt{3}, we get: [= frac{sqrt{3}(frac{sqrt{3}}{2}cos 10^{circ}-frac{1}{2}sin 10^{circ})}{cos 40^{circ}}.] Recognizing the expression inside the parentheses as cos 40^{circ}, we finally have: [= frac{sqrt{3}cos 40^{circ}}{cos 40^{circ}} = sqrt{3}.] Therefore, the answer is boxed{D}.

answer:1. **Base Case:** - Consider the base case where the number of vertices (imons) is 1. If there is only one imon, it cannot be entangled with any other imon. Therefore, the base case is trivially true. 2. **Inductive Hypothesis:** - Assume that for any graph with ( n-1 ) vertices, it is possible to apply a sequence of operations such that no two imons are entangled. 3. **Inductive Step:** - Consider a graph ( G ) with ( n ) vertices. We need to show that we can apply a sequence of operations to make sure no two imons are entangled. 4. **Odd Degree Vertex:** - If there exists a vertex ( v ) in ( G ) with an odd degree, we can destroy it using operation (i). This reduces the problem to a graph with ( n-1 ) vertices, where we can apply the inductive hypothesis. 5. **All Vertices Have Even Degree:** - If all vertices in ( G ) have even degrees, we proceed as follows: - Create a new graph ( G mathop{square} K_2 ), which is the Cartesian product of ( G ) and ( K_2 ). This means we create a copy ( G' ) of ( G ) and for each vertex ( v ) in ( G ), there is a corresponding vertex ( v' ) in ( G' ). Each vertex ( v ) in ( G ) is connected to its copy ( v' ) in ( G' ). 6. **Degree in ( G mathop{square} K_2 ):** - In ( G mathop{square} K_2 ), each vertex ( v ) in ( G ) has an odd degree because it is connected to its copy ( v' ) in ( G' ), and vice versa. Therefore, every vertex in ( G mathop{square} K_2 ) has an odd degree. 7. **Destroying Odd Degree Vertices:** - We can now destroy all vertices with odd degrees in ( G mathop{square} K_2 ) using operation (i). This will eventually leave us with a graph where no vertices are entangled. 8. **Handling Remaining Vertices:** - If after destroying vertices, we are left with an empty set ( H ), we are done. If not, we undo the last deletion and instead delete the copy ( v' ) of the vertex ( v ). This ensures that ( v ) becomes isolated and can be ignored. 9. **Algorithm for Good Vertices:** - Define a vertex ( v ) as "good" if ( deg v + deg v' ) is odd. Since all vertices in ( H ) are good, we can perform the following: - Choose a good vertex ( v ) in ( H ). If ( deg v ) is even, erase ( v' ); otherwise, erase ( v ). - This process ensures that the degrees of remaining vertices are adjusted correctly, and the algorithm continues until ( H ) becomes empty. 10. **Applying Inductive Hypothesis:** - Once ( H ) is empty, we apply the inductive hypothesis to the remaining graph, ensuring no two imons are entangled. Thus, by induction, we have shown that it is possible to apply a sequence of operations to ensure that no two imons are entangled. (blacksquare)

answer:(1) When a = 1, the function becomes f(x) = |3x - 1| + x + 3. For x geq frac{1}{3}, the inequality f(x) leq 5 can be rewritten as 3x - 1 + x + 3 leq 5, which simplifies to frac{1}{3} leq x leq frac{3}{4}. For x < frac{1}{3}, the inequality f(x) leq 5 can be rewritten as -3x + 1 + x + 3 leq 5, which simplifies to -frac{1}{2} leq x < frac{1}{3}. Thus, the solution set for the original inequality is boxed{-frac{1}{2} leq x leq frac{3}{4}}. (2) The function f(x) = |3x - 1| + ax + 3 can be represented as a piecewise function: f(x) = begin{cases} (3 + a)x + 2, & x geq frac{1}{3} (a - 3)x + 4, & x < frac{1}{3} end{cases} For the function f(x) to have a minimum value, it is necessary and sufficient that the following conditions hold: begin{cases} 3 + a geq 0 a - 3 leq 0 end{cases} This simplifies to -3 leq a leq 3. Therefore, the range of real number a is boxed{-3 leq a leq 3}.