1.已知x为整数,且2/(x+3)+2/(3-x)+2x+18/x²-9为整数,求所有符合条件的x的值2.解方程2(x²+1/x²)-3(x+1/x)-1=03.试证：对任意的正整数n,有1/1*2*3 + 1/2*3*4 + ...+1/n(n+1)(n+2)

1、那个式子是不是 [2/(x+3)]+[2/(3-x)]+(2x)+[18/(x²-9)]？
或者是 [2/(x+3)]+[2/(3-x)]+[(2x+18)/(x²-9)]？
做法是先通分再讨论。
2、设 t=x+1/x，
则 x²+1/x²=(x+1/x)²-2=t²-2
所以，原方程化为
2t²-4-3t-1=2t²-3t-5=(2t-5)(t+1)=0
解得 t=5/2 或 t=-1
即 x=2、x=1/2 或 无解
3、裂项相消法
1/[n(n+1)(n+2)]=(1/2){1/[n(n+1)]-1/[(n+1)(n+2)]}
所以
不等式左边
=(1/2)(1/2-1/6)+(1/2)(1/6-1/12)+…+(1/2){1/[n(n+1)]-1/[(n+1)(n+2)]}
=(1/2){1/2-1/6+1/6-1/12+…+1/[n(n+1)]-1/[(n+1)(n+2)]}
=(1/2){1/2-1/[(n+1)(n+2)]}
=1/4-1/[2(n+1)(n+2)]
＜1/4

1.2/(x+3)+2/(3-x)+2x+18/x²-9=[2*(x-3)-2*(x+3)+18]/(x-3)(x+3)+2x=2x
是不是题目括号没加对，x是整数时，后面的式子也是整数
2.x²+1/x²=(x+1/x)*(x+1/x)-2
所以，另t=(x+1/x)，则原式=2*(t*t-2)-3t-1=2t*t-3*t-5=0
解出，t=-1,5/2
x+1/x=-1无解
x+1/x=5/2得出x=2,1/2

（1） 2/x+3+2/3-x+2x+18/x^2-9
=[2(X-3)-2(X+3)+2X+18]/(X+3)(X-3)
=2(X+3)/(X+3)(X-3)
=2/(X-3)
∴X-3=±1, ±2
∴X=4, 2, 5, 1
符合条件x的值的和为4+2+5+1=12
（2）令a=x+1/x
两边平方
a²=x²+2+1/x²
所以x²+1/x²=a²-2
所以2(a²-2)-3a-1=0
2a²-3a-5=0
(2a-5)(a+1)=0
a=5/2,a=-1
x+1/x=5/2
2x²-5x+2=0
(2x-1)(x-2)=0
x=1/2,x=2
x+1/x=-1
x²+x+1=0
此方程无解
所以x=1/2,x=2
3）1/n(n+1)(n+2)=1/n·[1/(n+1) - 1/(n+2)]=1/n(n+1) - 1/n(n+2) =[1/n-1/(n+1)] - ½[1/n-1-(n+2)]
=½[1/n-2/(n+1)+1/(n+2)].
∴原式=½(1/1-2/2+1/3)+½(1/2-2/3+1/4)+½(1/3-2/4+1/5)+···+½[1/n-2/(n+1)+1/(n+2)]
=½(1/1+1/2+1/3+···+1/n) - 1·[1/2+1/3+1/4+···+1/(n+1)] + ½[1/3+1/4+···+1/(n+2)]
=½[1+1/2+1/(n+1)+1/(n+2)] + 1·[1/3+1/4+···+1/n] - 1·[1/2+1/3+1/4+···+1/(n+1)]
=1/2+1/4 + ½[1/(n+1)+1/(n+2)] - [1/2+1/(n+1)]
=1/4-½[1/(n+1)-1/(n+2)]

好，揽下。做会