答:
原方程即:
x[1/(1*2)+1/(2*3)+...+1/(2008*2009)]=2008
x[(1/1-1/2)+(1/2-1/3)+...+(1/2008-1/2009)]=2008
x(1-1/2009)=2008
x*2008/2009=2008
x=2009
答:
原方程即:
x[1/(1*2)+1/(2*3)+...+1/(2008*2009)]=2008
x[(1/1-1/2)+(1/2-1/3)+...+(1/2008-1/2009)]=2008
x(1-1/2009)=2008
x*2008/2009=2008
x=2009