计算:(x+y)(x-2y)-my^2(m,n均为常数)的值.在把x,y的值代入计算时..计算:(x+y)(x-2y)-my^2(m,n均为常数)的值.在把x,y的值代入计算时粗心的小名和小亮都把y的值看错,但结果都等于25.细心的小
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![计算:(x+y)(x-2y)-my^2(m,n均为常数)的值.在把x,y的值代入计算时..计算:(x+y)(x-2y)-my^2(m,n均为常数)的值.在把x,y的值代入计算时粗心的小名和小亮都把y的值看错,但结果都等于25.细心的小](/uploads/image/z/9299596-4-6.jpg?t=%E8%AE%A1%E7%AE%97%EF%BC%9A%EF%BC%88x%2By%EF%BC%89%28x-2y%29-my%5E2%28m%2Cn%E5%9D%87%E4%B8%BA%E5%B8%B8%E6%95%B0%EF%BC%89%E7%9A%84%E5%80%BC.%E5%9C%A8%E6%8A%8Ax%2Cy%E7%9A%84%E5%80%BC%E4%BB%A3%E5%85%A5%E8%AE%A1%E7%AE%97%E6%97%B6..%E8%AE%A1%E7%AE%97%EF%BC%9A%EF%BC%88x%2By%EF%BC%89%28x-2y%29-my%5E2%28m%2Cn%E5%9D%87%E4%B8%BA%E5%B8%B8%E6%95%B0%EF%BC%89%E7%9A%84%E5%80%BC.%E5%9C%A8%E6%8A%8Ax%2Cy%E7%9A%84%E5%80%BC%E4%BB%A3%E5%85%A5%E8%AE%A1%E7%AE%97%E6%97%B6%E7%B2%97%E5%BF%83%E7%9A%84%E5%B0%8F%E5%90%8D%E5%92%8C%E5%B0%8F%E4%BA%AE%E9%83%BD%E6%8A%8Ay%E7%9A%84%E5%80%BC%E7%9C%8B%E9%94%99%2C%E4%BD%86%E7%BB%93%E6%9E%9C%E9%83%BD%E7%AD%89%E4%BA%8E25.%E7%BB%86%E5%BF%83%E7%9A%84%E5%B0%8F)
计算:(x+y)(x-2y)-my^2(m,n均为常数)的值.在把x,y的值代入计算时..计算:(x+y)(x-2y)-my^2(m,n均为常数)的值.在把x,y的值代入计算时粗心的小名和小亮都把y的值看错,但结果都等于25.细心的小
计算:(x+y)(x-2y)-my^2(m,n均为常数)的值.在把x,y的值代入计算时..
计算:(x+y)(x-2y)-my^2(m,n均为常数)的值.在把x,y的值代入计算时粗心的小名和小亮都把y的值看错,但结果都等于25.细心的小红把正确的x,y的值代入计算,结果恰好也是25.为了探个究竟,她又把y的值随机换成2009,结果仍为25.
(1)根据上面的情况,试探究其中的奥妙
(2)你能确定m,
计算:(x+y)(x-2y)-my^2(m,n均为常数)的值.在把x,y的值代入计算时..计算:(x+y)(x-2y)-my^2(m,n均为常数)的值.在把x,y的值代入计算时粗心的小名和小亮都把y的值看错,但结果都等于25.细心的小
请仔细看看你的题,是不是有错误.n在何处?
请问哪有n呀,
此类题可以转化为(x+y)(x-2y)-my^2=25对于任意Y值都成立的恒成立问题
可以通过取特殊值或化成关于Y的二次方程求解
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