解方程:1⼀x(x+2)+1⼀(x+1)(x+3)+…+1⼀(x+8)(x+10)=0

急求!!过程要清楚!谢谢各位亲们
2025-06-28 07:03:51
推荐回答(2个)
回答1:

1/2*[1/x-1/(x+2)]+1/2*[1/(x+1)-1/(x+3)]+…+1/2*[1/(x+8)-1/(x+10)]=0
1/2*[1/x-1/(x+2)+1/(x+1)-1/(x+3)+1/(x+2)-1/(x+4)+…+1/(x+8)-1/(x+10)]=0
1/2*[1/x+1/(x+1)-1/(x+9)-1/(x+10)]=0
所以1/x+1/(x+1)-1/(x+9)-1/(x+10)=0
[1/x-1/(x+9)]+[1/(x+1)-1/(x+10)]=0
9/[x(x+9)]+9/[(x+1)(x+10)]=0
9/[x(x+9)]=-9/[(x+1)(x+10)]
所以x(x+9)=-(x+1)(x+10)
x^2+9x=-(x^2+11x+10)
x^2+9x+x^2+11x+10=0
2x^2+20x+10=0
x^2+10x+5=0
x=-5±2√5

打字不易,望采纳

回答2:

楼上是对的,把1/x(x+2)拆成1/2*[1/x-1/(x+2)],
1/(x+1)(x+3)拆成1/2*[1/(x+1)-1/(x+3)]
。。。。。
然后计算就好了