Question

lim(3x-根号下(ax^2+bx+c))=2,求abc

Answer 1

x-->1

lim(3x-根号下(ax^2+bx+c))

=lim[9x^2-(ax^2+bx+c)]/(3x+根号下(ax^2+bx+c))

=2

则 9-a=0,a=9

lim（-bx-c)/(3x+根号下(9x^2+bx+c))

=lim （-b-c/x)/(3+根号下(9+b/x+c/x^2))

=lim (-b)/6

=2

b=-12

c为任意实常数