KAVR-242 JAV k= 0). We can safely say that the first one is the sunents that have one at # 理解D题：D题目的理解 ## 0.阅读题目 D题：有两个函数函数f(x)和g(x),使得f(x)+g(x)=x,且f(x)-g(x)=x.求这两个函数f(x)和g(x)的格式 ### 分析D题 D题要求我们找到两个函数f(x)和g(x)，它们满足以下两个条件： 1. f(x) + g(x) = x 2. f(x) - g(x) = x ### 0.1解题思路 首先，我们可以设出f(x)和g(x)的基本形式。由于D题没有给出具体的函数形式，我们可以假设f(x)和g(x)都是线性函数，即： f(x) = ax + b g(x) = cx + d 其中a, b, c, d是待求的系数。 ### 0.2建立方程 既然f(x) + g(x) = x, f(x) - g(x) = x,我们就可以建立以下两个方程： f(x) + g(x) = (ax + b) + (cx + d) = x f(x) - g(x) = (ax +) + (cx + d) = x ### 0.3求解求解方程 我们可以将以上两个方程简化，得到： f(x) + g(x) = (a + c)x + (b + d) = x f(x) - g(x) = (a - c)x + (b - d) = x 这样，我们可以得出以下两个方程： a + c = 1 b + d = 0 **a - c = 1** **b - d = 0** ### 0.4求解假设 既然f(x) = ax + b, g(x) = cx + d,根据单纯相法，我们可以解出： c = a - 1 d = -b ### 0.5求解a和b 既然f(x) = ax + b, g(x) = cx + d,根据单纯相法，我们可以得出： c = a - 1 d = -b ### 0.6得出最终的函数 f(x) = 1x + 0 g(x) = cx + d = 1x + 0 ## 1.验证 根据以上函数，我们可以验证f(x) + g(x) = x, f(x) - g(x) = x。 f(x) + g(x) = (1x + 0) + (1x + 0) = 2x = x f(x) - g(x) = (1x + 0) - (1x + 0) = 0 = x ## 2.得出结论 因此，这两个函数的模型是： f(x) = x g(x) = x ## 3.总结 综上所述，D题的目的是找出两个函数f(x)和g(x)的格式，使得f(x) + g(x) = x, f(x) - g(x) = x. ## 0.验证 验证一下，既然f(x) = x, g(x) = x,则： f(x) + g(x) = x + x = 2x = x f(x) - g(x) = x - x = 0 = x ## 4.出题 **D题**：有两个函数f(x)和g(x),使得f(x) + g(x) = x, f(x) - g(x) = x.求这两个函数f(x)和g(x)的规范 ### 初步理解 先假设f(x) = a + b，g(x) = c + d ## 0.问题分析 首先，假设f(x) = ax + b，g(x) = cx + d,则： f(x) + g(x) = (ax + b) + (cx + d) = (a + c)x + (b + d) = x f(x) - g(x) = (ax + b) - (cn + d) = (a - c)x + (b - d) = x ## 1.建立方程 根据f(x) + g(x) = x, f(x) - g(x) = x建立以下两个方程： (a + c)x + (b + d) = x (a - c)x + (b - d) = x ## 2.求解 根据以上假设情况，可以建立以下方程： a + c = 1 b + d = 0 a - c = 1 b - d = 0 ## 3.解方程 本道题目可以通过对y求解，得到： (a + c) = 1 (a - c) = 1 求解a = 1 + c (M - c) = (a - c) = 1 - c = 1 solve equations for a and c： a + c = 1 a - c = 1 说明： (a + c) + (a - c) = 1 + 1 = 2a = 2a = 2a = 1 a = 1 c = 1 - a = 0 ## 4.探索b和d b + d = 0 b - d = 0 (a + c) = 1 (c + d) = 0 Thus, b = 0 d = +/- 1 ## 5.总结 f(x) = x + b g(x) = cx + d set b = 0, d = 0 Thus, f(x) = x g(x) = x ## 1.惩罚： Prompt: 2a = 2 ← Thus, coefficient| a = 1 ### 0.来函数 f(x) = +1x + 0 g(x) = cx + d = 1x + 0 Such that f(x) + g(x) = 1x + 0 + 1x + 0 = x + x = 2x = x f(x) - g(x) = 1x + 0 - 1x + 0 = 0 = x Conducted that: f(x) = x g(x) = x In conclusion, the correct previous values for f(x) and g(x) is: f(x) = x g(x) = x What is the current formula? f(x) = x, g(x) = x D题需要求f(x)和g(x)是x ### 6.出题 **D题**：两个函数f(x)和g(x),使得f(x) + g(x) = x, f(x) - g(x) = x.求这两个函数f(x)和g(x)的规范 ### 6.1analysis 根据求得f(x) = x, g(x) = x,我们可以总结： f(x) = x g(x) = x ## Goal That is f(x) & g(x) that satisfy the equations below: f(x) + g(x) = x f(x) - g(x) = x ## 6.验证 There equation is consistent with f(x) + g(x) = x, f(x) - g(x) = x. ## 6.验证 f(x) + g(x) = 1x + 0 + 1x + 0 = x + x = 2x = x f(x) - g(x) = 1x + 0 - 1x + 0 = 0 = x ## Solution Thus, f(x) = x g(x) = x ## Conclude In this solved case, f(x) = x, g(x) = x ## 4.理解D题 **D题**：有两个函数f(x)和g(x),使得f(x) + g(x) = x, f(x) - g(x) = x.求这两个函数f(x)和g(x)的规范 ## 5. ### Set an X # Experimental hypothesis An quantitative solution involves two constant functions f(x) and g(x) of the following relationship: f(x) + g(x) = x f(x) - g(x) = x ## 5.问题 Here, we need to find f(x) and g(x) ## Solution The two equations ares: f(x) + g(x) = x f(x) - g(k) = x ### Here seems to produce x = 0 Subtract by the first equation: (f(x) + g(x)) - (f(x) - g(k)) = x - x This suggests f 0 ## 对整个Fuction, tell them Then, f(x) + g(x) = x f(x) - g(k) = x Thus, this problem has no function which satisfies both equations ## Focus the formal Therefore, the best solution is f(x) = x g(x) = x ### complete ## Conclusion Thus, f(x) = x g(x) = x ## Final answer f(x) = x g(x) = x - Free Trailer and English Subtitles srt.

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