6y^2-3y-4=0两边除以-y^2-6+3/y+4/y^2=04*(1/y)^2+3*(1/y)-6=0又4x^2+3x-6=0 所以x和1/y是方程4a^2+3a-6=0的两个根所以由韦达定理x+1/y=-3/4,x*1/y=-3/2所以(2xy+5x+2)/3y =2x/3+5x/3y+2/3y=（2/3)(x+1/y)+(5/3)*x*1...