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什么叫方程的解?什么叫做解方程
纸弟弟烧饼60
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使方程左右两边相等的未知数的值叫做方程的解叫方程的解求方程中的未知数,叫做解方程
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代入后使方程成立的数解方程：求方程的解的过程
含有未知数的等式叫做方程，使方程左右两边相等的未知数的值叫做方程的解，求方程解的过程叫做解方程。
方程的解是结果，解方程是过程！
扫描下载二维码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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