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高二第二試题目解法 1.証明:不論n是什么整数,方程 x~2-16nx+7~5=0 (1)没有整数解。这题目里面的7~5可以改成7~8,其中s是任何正的奇数。解题时,最好利用根与系数的关系,并用反证法。现在把解写在下面: 解:设两根为x_1,x_2,则有 x_1+x_2=16n (2) x_1x_2=7~8 (3)现在假定(1)有一根是整数,则由(2),另一根也是整数。因7是素数,故由(3)知,x_1x_2可以写成下面的形式: x_1=±7~k,x_2=±7~h (4)上面两式同时取+号或-号,而 k+h=s. (5)把(4)代入(2)得 7~k+7~h=±16n (6)因k+h=s为奇数,不妨设k>h,则
High school second test title solution 1. Proof: No matter what integer n is, the equation x~2-16nx+7~5=0 (1) There is no integer solution. 7~5 in this topic can be changed to 7~8, where s is any positive odd number. When solving a problem, it is better to use the relationship between the root and the coefficient and use the counter-evidence method. Now write the solution below: Solution: Let two are x_1, x_2, then there are x_1+x_2=16n (2) x_1x_2=7~8 (3) Now suppose that (1) has one integer, then by (2) The other root is also an integer. Since 7 is a prime number, it is known from (3) that x_1x_2 can be written in the following form: x_1=±7~k,x_2=±7~h (4) The above two equations take the + or - sign at the same time, and k+h =s. (5) Substitute (4) into (2) to get 7~k+7~h=±16n (6) Since k+h=s is an odd number, we may set k>h.