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

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=0

3.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)

问答/363℃/2024-05-01 13:47:21

优质解答：

（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)]

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