若|x-4|=根号3y+1=0求[(3x+1)²+(3x+y)(3x-y)+2xy]÷(-2xy)的值

x=4,y=-1/3
[(3x+1)²+(3x+y)(3x-y)+2xy]÷(-2xy)
=[(3x+1)²+(9x^2-y^2)+2xy]÷(-2xy)
=[(9x^2+6x+1)+(9x^2-y^2)+2xy]÷(-2xy)
=-[18x^2+6x+1-y^2+2xy]÷(2xy)
=-9x/y-3/y-1/(2xy)+y/(2x)-1
=108+9+3/4-1/24-1
=116+18/24