Question

解方程1/x^2+x+1/x^2+3x+2+1/x^2+5x+6+.+1/x^2+199x+9900=100/x^2+1

Answer 1

1/x^2+x+1/x^2+3x+2+1/x^2+5x+6+.+1/x^2+199x+9900

=1/x(x+1)+1/[(x+1)(x+2)]++1/[(x+2)(x+3)]+……+1/[(x+99)(x+100)]

=1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+……+1/(x+99)-1/(x+100)=1/x-1/(x+100)

所以：1/x-1/(x+100)=100/(x^2+100)

所以：(x+100)(x^2+100)-x(x2+100)=100x(x2+100)

化简后得到：10000x=10000

所以：x=1