英语解数学题的格式是什么？
英语解数学题的格式是什么？
09-01-12 &
Problem: [BRITAIN 1984/4] x^2 - |_x^2_| = ( x - |_x_| )^2. Count the number of solutions 1&=x&=n. Solution: Let x = m + d, where m = |_x_| and d = x - |_x_| which is 0&=d&1. Plugging x = m + d into x^2 - |_x^2_| = ( x - |_x_| )^2 we get (m + d)^2 - |_(m + d)^2_| = d^2 and after simplification 2md = |_2md + d^2_|. Therefore 2md is an integer k. Case m = 0: Then every 0&=d&1 is possible. Therefore all 0 &= x & 1 are valid. Case m & 0: Then 2md = k. That is d = k/(2m). As 0&=d&1 we have 0 &= k/(2m) & 1. So possible values are k in {0, 1, 2, ..., 2m-1 }. The corresponding x values are m, m + 1/(2m), m + 2/(2m), ..., m + (2m-1)/(2m). Let L(a,b) = { a&=x&b : x^2 - |_x^2_| = ( x - |_x_| )^2 }. Then we have shown #L(n,n+1) = 2n for positive integers n. Summing up gives #L(1,n) = n(n-1). The number of solutions 1&=x&=n is n(n-1) + 1. Of these are n integral and (n-1)^2 non-integral. Problem: Find all positive integers a and b for which |_a^2/b_| + |_b^2/a_| = |_(a^2+b^2)/(ab)_| + ab. Problem: If the following is true for all integers n |_n/a_| = |_n/b_| must a = b ? a and b are real numbers. Solution: |_n/a_| = |_n/b_| =& | n/a - n/b | & 1 =& | 1/a - 1/b | & 1/n Since this is true for all n, 1/a = 1/b. Problem: [Jan87, EdM P964] Find all real pairs (a,b) such that for all positive integers n a |_b n_| = b |_a n_|. Solution: [BrF88, EdM P964] It is clear that a |_b n_| = b |_a n_| for all natural numbers n if either ab = 0, or if a=b, or if a and b are both integers. We show that this condition is also necessary. Thus we suppose a |_b n_| = b |_a n_| for all n, ab != 0, and a != b. Then, taking n = 1, we have bm = ak, where m = |_a_| and k = |_b_|. Thus 2m &= 2a & 2m + 2, so that either 2m &= 2a & 2m + 1 or 2m + 1 &= 2a & 2m + 2. Similarly, either 2k &= 2b & 2k + 1 or 2k + 1 &= 2b & 2k + 2. Taking n = 2 we conclude that in fact |_2a_| = 2m and |_2b_| = 2k. (Each of the other possibilities contradicts one of our hypotheses. E. g. assume |_2a_| = 2m + 1 and |_2b_| = 2k + 1 then b(2m+1) = a(2k+1) and as bm = ak we have the contradiction b = a.) Repeating this argument we inductively establish that |_2^r a_| = 2^r m and |_2^r b_| = 2^r k, so that m &= a & m + 1/2^r and k &= b & k + 1/2^r for all natural numbers r. Thus a = m and b = k, and our asertion is proven. Problem: (Sillke) Find all real pairs (x,y) such that x |_y_| = y |_x_|. Problem: Komal F3232 Let b(n) denote the minimum value of expression k + n/k, where k is a positive integer. Prove that for any natural number n, |_b(n)_| = |_sqrt(4n+1)_| Problem: Komal Gy2047 Solve the equation |_sqrt(|_x_|)_| = |_sqrt(sqrt(x))_| on the set of real numbers. Solution: First observe that |_sqrt(|_x_|)_| = |_sqrt(x)_|. Second set x = z^4 and get the nicer looking equation |_z^2_| = |_z_|. Case z & 0: |_z_| &= z & 0 &= |_z^2_|. No solution. Case z &= tau = (sqrt(5) + 1)/2: As z^2 &= tau*z &= z + 1 we get |_z^2_| &= |_z + 1_| = |_z_| + 1 & |_z_|. No solution. Case 0 &= z & tau: Analyze the range of |_z_| which is {0, 1}. So there are only two cases left. Case |_z^2_| = 0 = |_z_|: solutions for 0 &= z & 1. Case |_z^2_| = 1 = |_z_|: solutions for 1 &= z & sqrt(2). Collecting the results we get the solution 0 &= x & 4. Problem: Ouardini Problem 2-10 Solve the equation |_x^(1/2)_| = |_x^(1/3)_| on the set of real numbers. Solution: First set x = z^6 and get the nicer looking equation |_z^3_| = |_z^2_|. Case z & 0: |_z^3_| &= z^3 & 0 &= |_z^2_|. No solution. Case z &= 3/2: As z^3 &= 3/2 z^2 &= z^2 + 1 we get |_z^3_| &= |_z^2 + 1_| = |_z^2_| + 1 & |_z^2_|. No solution. Case 0 &= z & 3/2: Analyze the range of |_z^2_| which is {0, 1, 2}. So there are only three cases left. Case |_z^3_| = 0 = |_z^2_|: solutions for 0 &= z & 1. Case |_z^3_| = 1 = |_z^2_|: solutions for 1 &= z & 2^(1/3). Case |_z^3_| = 2 = |_z^2_|: solutions for 2^(1/2) &= z & 3^(1/3). Collecting the results we get the solution 0 &= x & 2 and 8 &= x & 9. Problem: SSM 3696 Solve the equation |_sqrt(x)_| = |_x/k_| on the set of real numbers, where k is an integer. Problem: 20th MMO 9.1.2 Solve the equation x^3 - |_x_| = 3 on the set of real numbers. Solution: Rearranging the equation gives x^3 = 3 + |_x_|. Therefore the right hand side is an integer. Case x&=2: x^2 &= 4 =& x^3 &= 4x &= 3 + x &= 3 + |_x_|. No solution. Case 2&x&=1: x^3 = 3 + |_x_| = 4 =& x = 4^(1/3) = 1.587401 Case 1&x&=0: x^3 = 3 + |_x_| = 3 =& x & 1. No solution. Case 0&x&=-1: x^3 = 3 + |_x_| = 2 =& x & 1. No solution. Case x & -1: x^2 &= 1 =& x^3 & x & 2 + x & 3 + |_x_|. No solution. Problem: Ouardini Problem 2-12 Solve the equation 1 + sin^2(x) + sin^2(x - |_sqrt(x)_|) = cos(x) on the set of real numbers. Solution: This equation looks difficult by the obeservation 1 + sin^2(x) + sin^2(x - |_sqrt(x)_|) &= 1 &= cos(x) makes it rather simple. So we are looking for the common solution of the three equations: cos(x) = 1, sin(x) = 0, and sin(x - |_sqrt(x)_|) = 0. The first has the solutions 2Pi*Z, the second Pi*Z. So we have only to check x = x_n = 2Pi*n with n in Z. But x_m = x_n - k for an integer k has only one solution k = 0 as Pi is irrational. As k=0 means x=0 we have only one solution for the original equation. Problem: Komal C596 Solve the equation |_1/(1-x)_| = |_1/(1.5-x)_| on the set of real numbers. Solution: case x & 1: 0 & 1-x & 1.5-x k &= 1/(1.5-x) & 1/(1-x) & k+1 k=0: x&0 case x & 1.5: 1-x & 1.5-x & 0 k=-1: x&=2.5 Problem: Komal C605 (Dec 2000) Solve the equation 1 1 ----- + --------- = x |_x_| x - |_x_| Solution: Multiply the equation by |_x_| and x - |_x_|. This gives x = x |_x_| (x - |_x_|) Case x=0: This don't solve the original equation. Case x!=0: Cancel x 1 = |_x_| (x - |_x_|) =& x = |_x_| + 1/|_x_|. So for each integer n&=2 we get a valid solution x = n + 1/n. Problem: Determine the number of real solutions a of the equation |_a/2_| + |_a/3_| + |_a/5_| = a. Solution: There are 30 solutions. Since |_a/2_|, |_a/3_|, and |_a/5_| are integers, so is a. Now write a = 30p + q for integers p and q, 0 &= q & 30. Then |_a/2_| + |_a/3_| + |_a/5_| = a &=& 31p + |_q/2_| + |_q/3_| + |_q/5_| = 30p + q &=& p = q - |_q/2_| - |_q/3_| - |_q/5_|. Thus, for each value of q, there is exactly one value of p (and one value of a) satisfying the equation. Since q can equal any of thirty values, there are exactly 30 solutions as claimed. 参考资料:
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数学题求解?60&x00B2;=X²，X=?
求大神指教
这道题是：每后面一个数...
求 x的是多少，图中横列...
我就不相信没人解不出来...
求解一道数学题： 王小...
五题不会做
不能用x方程的
给跪了。听说是印度阿三...
已知本人还有100大洋，...
到数学吧发吧
逗比！！！
用根号吧。骚年
你确定是60度角
开不完，30又根号3
这。。。。。。X=30根号3
1：根号3：2
1 2 根号三 打倍数就出来了惹　　——我是新人什么都不懂你们不要欺负我惹！
60×cos30度＝3倍根号3
30√3初中生问题
答案又不一定是整数，30倍根号三
三十倍根号三
题有问题矛盾
逗比，你试试你能是画出这个三角形，还30度的，逗比
目测lz小学生
lz小学生，鉴定完毕。————————————————♚
直接开方x平方=2700 ，所以x=30根号3
回复 爱在风花雪 :题目没说是正的就要带上负的。三角形是楼猪意淫的吧
好像直接口算出来吧，楼主几年级？
屁大事也发帖
三角形不存在
百度小说人气榜
贴吧热议榜
使用签名档&&
保存至快速回贴求解数学题一种胶卷的价格是20元,共拍了36张,冲洗费为19.6元.小丽拍了14张,小方拍了22张.它们分别应付多少钱?望能够详细讲解
飞机の69鋼V
每张 （20+19.6）÷36＝39.6÷36＝1.1元小丽 14x1.1=15.4元小方 22x1.1=24.2元
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