求方程1/x+1/X+1+1/X+2+1/X+3=19/20的正整数解
问题描述：
求方程1/x+1/X+1+1/X+2+1/X+3=19/20的正整数解
下面这个过程我看不懂.
X为正整数,那么先看范围,左边小于4/x,大于4/(x+3),先解方程 4/x=19/20,得x=4.2
再解4/(x+3)=19/20,得x=1.2
则正整数解的范围 1.2 ≤ x≤4.2
x的可能值为2,3,4,每个都带进去算,最终确定解为x=3
答
因为分母越大,分数越小,所以x,x+1,x+2,x+3都大于等于x,都小于等于x+3所以分数1/x,1/(x+1),1/(x+2),1/(x+3)都小于等于1/x,都大于等于1/(x+3)把他们加起来1/(x+3)+1/(x+3)+1/(x+3)+1/(x+3)≤1/x+1/(x+1)+1/(x+2)+1/(x+...