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)
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/31 03:58:29
![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)](/uploads/image/z/667666-10-6.jpg?t=1.%E5%B7%B2%E7%9F%A5x%E4%B8%BA%E6%95%B4%E6%95%B0%2C%E4%B8%942%2F%28x%2B3%29%2B2%2F%283-x%29%2B2x%2B18%2Fx%26%23178%3B-9%E4%B8%BA%E6%95%B4%E6%95%B0%2C%E6%B1%82%E6%89%80%E6%9C%89%E7%AC%A6%E5%90%88%E6%9D%A1%E4%BB%B6%E7%9A%84x%E7%9A%84%E5%80%BC2.%E8%A7%A3%E6%96%B9%E7%A8%8B2%28x%26%23178%3B%2B1%2Fx%26%23178%3B%29-3%28x%2B1%2Fx%29-1%3D03.%E8%AF%95%E8%AF%81%EF%BC%9A%E5%AF%B9%E4%BB%BB%E6%84%8F%E7%9A%84%E6%AD%A3%E6%95%B4%E6%95%B0n%2C%E6%9C%891%2F1%2A2%2A3+%2B+1%2F2%2A3%2A4+%2B+...%2B1%2Fn%28n%2B1%29%28n%2B2%29)
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.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)
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) 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
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...
全部展开
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
收起
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<...
全部展开
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
收起