分式方程1/[x(x+1)]+1/[(x+1)(x+2)]+1/[(x+2)(x+3)]+...+1/[(x+9)(x+10)]=5/12

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分式方程1/[x(x+1)]+1/[(x+1)(x+2)]+1/[(x+2)(x+3)]+...+1/[(x+9)(x+10)]=5/12分式方程1/[x(x+1)]+1/[(x+1)(x+2)]+1

分式方程1/[x(x+1)]+1/[(x+1)(x+2)]+1/[(x+2)(x+3)]+...+1/[(x+9)(x+10)]=5/12
分式方程1/[x(x+1)]+1/[(x+1)(x+2)]+1/[(x+2)(x+3)]+...+1/[(x+9)(x+10)]=5/12

分式方程1/[x(x+1)]+1/[(x+1)(x+2)]+1/[(x+2)(x+3)]+...+1/[(x+9)(x+10)]=5/12
即1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+...+1/(x+9)-1/(x+10)=5/12
1/x-1/(x+10)=5/12
两边乘12x(x+10)
120=5x²+50x
x²+10x-24=0
(x+12)(x-2)=0
x=-12,x=2

1/(x+n)(x+n+1)=1/(x+n)-1/(x+n+1)
所以原式可化为1/x-1/(x+1)+1/(x+1)-1/(x+2)……-1/(x+10)=1/x-1/(x+10)=5/12
化简的x平方+10x-24=0可解x=2/-12
最后检验是否为增根。。