这道题怎么证明_百度知道
这道题怎么证明
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因为f(x)在x趋于无穷时的极限存在，所以存在一个N,
使得对于任意的充分小的e 只要x,y&=N, 就有|f(x)-A|&=e 和 |f(y)-A|&=e，这意味着|f(x)-f(y)|&=|f(x)-A|+|f(y)-A|&=2e。由假设知此处e不依赖于x,y的取值所以 f(x)在区间[N，∞)上是一致收敛的。其次，在闭区间[a,N]上连续函数 f(x)也是一致收敛的。最后，对于x&N，y&N这种情况，我们可以这样估计|f(y)-f(x)|&=|f(y)-f(N)|+|f(N)-f(x)|&一个充分小的数，当x,y非常靠近时。所以 f(x)在[a,∞)上都是一致收敛的。
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这道题怎么做？数学证明题收藏
证全等就行。
兄弟证全等
= =高中竞赛题啊。你们看清楚好不好。。我觉得是 用余弦定理
反证法可不可以？
（还有我那个物理问题啊前辈别忘了去看看。= = ）
目测余弦定理差不多啊。设ABC三边abc。AD=xcosA=(b方+c方-a方)/2bcDE方=x方+(b-x)方+2x(b-x)cosA=EF方=DF方然后处理一下估计可以解出a=b=c(解不出就硬说你解出了- -)
看错题目了...我错了
先证三个小三角形全等，然后边等，最后为正三角形
试试正弦定理，完了三角函数运算……
2个方程3个未知数，但是有一个目测能消掉……
互相帮助真有爱
你们都被骗了，这是骗经验呢，其他的贴吧也见过，一样的题目
都说了是先利用里面那个正三角形 和题目给的那个等式就得到了外面那个三角形的各边都是相等的嘛！
果然这样 知识学多了 就不知道怎么用了 这么一个简单的初中问题 还硬要用什么正玄余玄的 ！ 唉！
这可不是持才放狂哦！
因为AD=BF=EC所以BD=A=CF且角A=角B=角C所以三角形ADE,BDF,CEF全等所以DE=FD=EF所以三角形DEF是正三角形
证明三个小三角形全等就OK了。AD=BF=EC，AB=BC=BC。所以BD=CF=AD，边角边，全等。
高中竞赛题呀！
本人还只是初中......
IBm98年的题，自行百度吧，分5种情况讨论哦ps，我也不会
这题我做过俺老师讲过
唉，看了18、19、20我果断知道了什么叫装B不成反被艹
图形确定，无脑解析
We proceed by breaking into cases. Let the sides of DEF have length x, and the line segments AD, BE, and CF have length y. The form of our solution will depend on the value of y/x. We begin with some special cases. The following Diagram is referenced for Case 1 & 2.Case 1, when y/x = 2/SQR3 In this case the largest value possible for angle A would be 60° --when angle ADF = 90° (see diagram). Since AD = BE = CF, this holds true for triangles BED and CFE as well. In this case then, since angles A, B, and C must sum to 180° , they must all be 60°, and ABC equilateral. Case 2, when y/x & 2/ SQR3 In this case, as x grows smaller (or y larger, or both) Angle A only grows smaller (see diagram). Again, since this is occuring equally throughout triangle ABC (AD = BE = CF, and triangle DEF is equilateral), angles A, B, and C must be under 60°, which makes it very difficult for them to sum to 180°. So case 2 is not possible. Case 3, when y/x = 1, or when triangle ADF is isosceles. For convenience, lable the angles as follows: ADF = beta, FAD = A, DFA = gamma, CFE = delta, and BED = epsilon. We know that gamma = A = 90° - beta / 2. This implies that delta = 180° - (gamma + 60°). Substituting (90° - (beta / 2)) for gamma, we get delta = (beta / 2) + 30°. Similarly, epsilon = (delta / 2) + 30°, and beta = (epsilon / 2) + 30°. The system of linear equations has the unique solution beta = &A = gamma = delta = epsilon = 60°. in particular A = 60° , and the same argument implies that all three angles of the large triangle are 60° and it is therefore equilateral. Case 4, when 0 & (y/x) & 1 Consider fixed values of x, y with 0 & y & x. Concentrate first on the triangel DAF, where we label the angles as in case 3 above and two sides are fixed at FD = x, DA = y. Let beta range from 0° to 180° and notice what happens to A and gamma. As beta increases, AF increases (by the law of cosines), and A decreases from 180° to 0°. By the law of sines, gamma at first increases from 0°, but when A hits 90°, gamma starts decreasing (if plotted on a graph, gamma’s range would describe a parabola).So there is exactly one value of beta, call it beta 0, which makes A = 60°. Let the angles in this case be called A0 = 60° and gamma0 = 180° - beta 0 - A0. We would have delta0 = 180° - 60° - gamma0 = beta0. Then each angle of the large triangle is 60° and it is equilateral. Suppose that all three angles beta, delta, epsilon, are smaller than beta0. Then each of the three angles A, B, and C is larger than 60°, and so can’t sum to 180°. The only other possibility is when the largest of the three, say beta, is larger than beta0. Then A & A0 = 60° & 90° so that gamma & gamma0 , and delta = 180° - 60° - gamma is also larger than beta0 . By the same argument, epsilon & beta0 . Then each of the three angles A, B and C is smaller than 60° , and again they can’t sum to 180°. So, in this case, the only consistent conclusion is when beta = beta0 and the large triangle is equilateral. Case 5, when 1 & (y/x) & 2 / SQR3 This last case is somewhat easier than # 4 above. Consider a movable triangle as before, but remember that y & x . So as beta grows from 0° to 180°, we notice that gamma shrinks from 180° to 0°, and delta grows. This means that beta, delta, epsilon
if delta were larger than beta, than (by monotonicity) epsilon would be larger than delta, and beta would be larger still, which would yield a contradiction. If beta, delta, and epsilon are the same, then by side-angle-side, the three triangles ADF, CFE, and BED are congruent, so the outer angles A, B, and C are equal, and the large triangle is equilateral.
这种高中竞赛题确实很伤高四党的脑筋
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这道题怎么证明
我有更好的答案
反证法,假设limf≠0,则limf=k或∞(k是常数)设h(x)=g(x)/f(x)①当limf=p时limh=lim(g/f)=limg/limf=0/p=0,即h是无穷小∴1/h是∞,即lim(f/g)=∞,矛盾.②当limf=∞时limh=lim(g/f)=lim(g*1/f)=limg*lim(1/f)=0*0=0,即h是无穷小∴由①得矛盾假设不成立,limf=0
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