answer:Solution: ((1)) In (triangle ABC), since (acos C+ccos A=2bcos A), we have (sin Acos C+sin Ccos A=2sin Bcos A), thus (sin (A+C)=sin B=2sin Bcos A), since (sin Bneq 0), we get (cos A= frac{1}{2}), which leads to: (A= frac{pi}{3}). ((2)) Since (cos A= frac{1}{2}= frac{b^{2}+c^{2}-a^{2}}{2bc}), (b+c= sqrt{10}), and (a=2), we have (b^{2}+c^{2}=bc+4), which leads to: ((b+c)^{2}=3bc+4=10), thus (bc=2). Therefore, (S= frac{1}{2}bcsin A= frac{sqrt{3}}{2}).Thus, the answers are: ((1)) (boxed{A= frac{pi}{3}}) ((2)) (boxed{S= frac{sqrt{3}}{2}})

answer:Since the function is given as f(x) = 4 + log_{a}(x - 2), with (a > 0 and a neq 1), its graph passes through the fixed point P(3, 4).The initial side of angle alpha coincides with the positive semi-axis of x, the vertex is at the coordinate origin, and the terminal side passes through the fixed point P. Thus, we have x = 3 and y = 4.The distance between the origin and point P is r = |OP| = sqrt{(3-0)^2 + (4-0)^2} = 5.Now, we can find tan{alpha} using the definition of tangent:tan{alpha} = frac{y}{x} = frac{4}{3}.To find the value of the given expression, we can rewrite it in terms of tan{alpha}:frac{sin{alpha} + 2cos{alpha}}{sin{alpha} - cos{alpha}} = frac{frac{tan{alpha}}{sqrt{1 + tan^2{alpha}}} + 2frac{1}{sqrt{1 + tan^2{alpha}}}}{frac{tan{alpha}}{sqrt{1 + tan^2{alpha}}} - frac{1}{sqrt{1 + tan^2{alpha}}}}.Simplifying and substituting the value of tan{alpha} = frac{4}{3}, we get:frac{sin{alpha} + 2cos{alpha}}{sin{alpha} - cos{alpha}} = frac{frac{4}{3} + 2}{1 - frac{4}{3}} = boxed{10}.

answer:For the given problems, let's break down the solutions step by step: Problem 1:Calculate frac{3x}{2x-y}-frac{x+y}{2x-y}.Step 1: Combine the fractions since they have the same denominator.frac{3x}{2x-y}-frac{x+y}{2x-y} = frac{3x-(x+y)}{2x-y}Step 2: Simplify the numerator.frac{3x-(x+y)}{2x-y} = frac{3x-x-y}{2x-y}Step 3: Further simplify the numerator.frac{3x-x-y}{2x-y} = frac{2x-y}{2x-y}Step 4: Since the numerator and denominator are the same, the fraction simplifies to 1.frac{2x-y}{2x-y} = 1So, the final answer for the first problem is boxed{1}. Problem 2:Calculate frac{{x}^{2}-5x}{x+2}÷frac{x-5}{{x}^{2}-4}.Step 1: Factorize the numerator and denominator where possible.frac{x^2-5x}{x+2}÷frac{x-5}{x^2-4} = frac{x(x-5)}{x+2} cdot frac{(x+2)(x-2)}{x-5}Step 2: Cancel out the common factors (x-5) and (x+2).frac{x(x-5)}{x+2} cdot frac{(x+2)(x-2)}{x-5} = x(x-2)Step 3: Expand the expression.x(x-2) = x^2-2xThus, the final answer for the second problem is boxed{x^2-2x}.

answer:AnalysisThis question mainly tests the calculation of function values. The key to solving this question is to determine the periodicity of the function based on the properties of the function's odd and even nature.SolutionSince f(x+2) is an even function, we have f(-x+2)=f(x+2). Since f(x) is an odd function, we have f(-x+2)=-f(x-2), which implies f(x+2)=-f(x-2), or equivalently f(x+4)=-f(x). Therefore, f(x+8)=-f(x+4)=f(x), which means f(9)=f(1)=1, f(8)=f(0). Since f(x) is an odd function, we have f(0)=0, which means f(8)=f(0)=0. Therefore, f(8)+f(9)=1+0=1. Hence, the correct answer is boxed{C}.

answer:Solution (Method 1 - "Take all and subtract the unwanted"): mathrm{P}_{mathrm{G}}^{mathrm{s}}-mathrm{P}_{5}^{2}=100 (items).(Solution 2 - "Take only the wanted"): 5 mathrm{P}_{5}^{2}=100 (items).

answer:Answer: 0Exam Point: Smart Calculation (Grouping Method)Analysis: Group according to the sign pattern “+- - +” with four as a group, the result of each group is 0, so the final result is 0.