answer:This question tests knowledge on: Factorization **Analysis of the approach:** First, factor out the common factor, then try using formula method, cross multiplication method, etc. **Detailed solution process:** ax^2 + 2axy + ay^2 = a(x^2 + 2xy + y^2) = a(x + y)^2 **Review of the question:** The result of factorization should be thorough, with each term being in its simplest form. Thus, the factorized form is boxed{a(x + y)^2}.

answer:Given a convex quadrilateral (ABCD), where only (angle D) is an obtuse angle. We need to show that to partition this quadrilateral into (n) obtuse triangles meeting the condition that (n geq 4). Step 1: Identify that there is a point (E) inside the quadrilateral (ABCD) where tri-segmentation occurs. Assume there is an internal point (E) in quadrilateral (ABCD). Then (E) must be a vertex of some obtuse triangle (EFG). Step 2: Discuss the different cases based on the position of (EFG). 1. **Case where edge (FG) (an edge of triangle (EFG)) lies on the boundary of (ABCD) (For example, let (FG) be on (AB))**: - **Situation (a)**: - When (FG) lies on (AB), triangle (EFG) splits (ABCD) such that we need at least additional divisions of the remaining part into obtuse triangles. - The pentagon (AEB CD) cannot be divided into just two triangles; thus, (n geq 4). - **Situation (b)**: - If (FG) lies internally parallel to (AB), forming a hexagon (AFEBCD). - This situation also requires more than one additional internal point for achieving obtuse triangles. - Thus, (n geq 4). - **Situation (c)**: - If (E) itself is connected such that non-boundary (FG) introduces more complex regions leading to a heptagon for example (AFEGBDC). - More internal segmentations are necessary to form obtuse triangles; hence (n geq 4). 2. **Case where (FG) does not lie on any boundary of (ABCD)**: - The edges (EF, FG,) and (GE) must each belong to distinct obtuse triangles to correctly partition the quadrilateral. - Each interior segment contributes to the necessity for additional triangles implying more than three obtuse triangles. - Therefore, (n geq 4). Conclusion: Analyzing all potential segmentations, it is clear that obtaining (n) obtuse triangles while maintaining the convex property and including interior points necessitates that (boxed{n geq 4}).

answer:According to the problem, we have 0 leqslant (k-1)x-1 leqslant (x+1)ln x holds for x in [1,2e]. When x in [1,2e], the graph of the function f(x)=(k-1)x-1 is a line segment, thus, begin{cases} f(1) geqslant 0 f(2e) geqslant 0 end{cases}, solving this yields k geqslant 2. On the other hand, k-1 leqslant dfrac {(x+1)ln x+1}{x} holds for x in [1,2e]. Let m(x)= dfrac {(x+1)ln x+1}{x}=ln x+ dfrac {ln x}{x}+ dfrac {1}{x}, then m'(x)= dfrac {x-ln x}{x^{2}}. Since 1 leqslant x leqslant 2e, (x-ln x)'=1- dfrac {1}{x} geqslant 0, thus, the function x-ln x is increasing, hence x-ln x geqslant 1-ln 1 > 0, so m'(x) geqslant 0, and the function m(x) is increasing over [1,2e]. Therefore, k-1 leqslant [m(x)]_{min}=m(1)=1, which means k leqslant 2. In summary, k=2. Hence, the answer is: boxed{{2}}. Consider the two cases g(x) leqslant f(x) and f(x) leqslant h(x) over the interval [1,2e]. This problem examines the properties of functions, and constructing a monotonic function over the interval is key to solving this problem. It is considered a medium-level question.

answer:To solve the problem, we need to find all positive integers ( n ) for which ((k(n))! cdot text{lcm}(1, 2, ldots, n) > (n-1)!). Here, ( k(n) ) denotes the number of composite numbers that do not exceed ( n ). 1. **Understanding ( k(n) )**: - ( k(n) ) is the count of composite numbers less than or equal to ( n ). - Composite numbers are numbers that have more than two distinct positive divisors. 2. **Rewriting the inequality**: - The given inequality is ((k(n))! cdot text{lcm}(1, 2, ldots, n) > (n-1)!). - The least common multiple (LCM) of the first ( n ) natural numbers can be expressed in terms of their prime factorization: (text{lcm}(1, 2, ldots, n) = p_1^{e_1} times p_2^{e_2} times cdots times p_k^{e_k}), where ( p_i ) are primes and ( e_i ) are their respective exponents. 3. **Analyzing the inequality**: - We need to compare ((k(n))! cdot p_1^{e_1} times p_2^{e_2} times cdots times p_k^{e_k}) with ((n-1)!). - For large ( n ), ((n-1)!) grows very quickly, so we need to find ( n ) such that the product of the factorial of the number of composite numbers and the LCM of the first ( n ) numbers exceeds ((n-1)!). 4. **Considering ( n ) of the form ( p^k )**: - If ( n ) is not of the form ( p^k ) (where ( p ) is a prime and ( k ) is an integer), it is unlikely that the inequality will hold because the right-hand side ((n-1)!) will be too large. - Therefore, we focus on ( n ) of the form ( p^k ). 5. **Checking small values of ( n )**: - We need to check small values of ( n ) to see if they satisfy the inequality. - For ( n = 2 ): [ k(2) = 0 quad text{(no composite numbers less than or equal to 2)} ] [ (k(2))! cdot text{lcm}(1, 2) = 1 cdot 2 = 2 ] [ (2-1)! = 1 ] [ 2 > 1 quad text{(True)} ] - For ( n = 3 ): [ k(3) = 0 quad text{(no composite numbers less than or equal to 3)} ] [ (k(3))! cdot text{lcm}(1, 2, 3) = 1 cdot 6 = 6 ] [ (3-1)! = 2 ] [ 6 > 2 quad text{(True)} ] - For ( n = 4 ): [ k(4) = 1 quad text{(4 is the only composite number less than or equal to 4)} ] [ (k(4))! cdot text{lcm}(1, 2, 3, 4) = 1 cdot 12 = 12 ] [ (4-1)! = 6 ] [ 12 > 6 quad text{(True)} ] - For ( n = 5 ): [ k(5) = 1 quad text{(4 is the only composite number less than or equal to 5)} ] [ (k(5))! cdot text{lcm}(1, 2, 3, 4, 5) = 1 cdot 60 = 60 ] [ (5-1)! = 24 ] [ 60 > 24 quad text{(True)} ] - For ( n = 7 ): [ k(7) = 2 quad text{(4 and 6 are the composite numbers less than or equal to 7)} ] [ (k(7))! cdot text{lcm}(1, 2, 3, 4, 5, 6, 7) = 2! cdot 420 = 2 cdot 420 = 840 ] [ (7-1)! = 720 ] [ 840 > 720 quad text{(True)} ] - For ( n = 9 ): [ k(9) = 4 quad text{(4, 6, 8, and 9 are the composite numbers less than or equal to 9)} ] [ (k(9))! cdot text{lcm}(1, 2, 3, 4, 5, 6, 7, 8, 9) = 4! cdot 2520 = 24 cdot 2520 = 60480 ] [ (9-1)! = 40320 ] [ 60480 > 40320 quad text{(True)} ] 6. **Conclusion**: - The only values of ( n ) that satisfy the inequality are ( 2, 3, 4, 5, 7, ) and ( 9 ). The final answer is ( boxed{ {2, 3, 4, 5, 7, 9} } )