#(2x-1)/(x+2)-1/x<=1#
#x/x*(2x-1)/(x+2)-(x+2)/(x+2)*1/x<=1#
#((2x^2-x)-(x+2))/(x(x+2))<=1#
#(2x^2-2x-2)/(x(x+2))<=1#
At this point, we must look at three cases. One where x is positive, and one where x is between -2 and 0, and one where x is less than -3. This is because the denominator reads #x(x+2)#. As you may know, multiplying an inequality by a negative value changes the direction of the inequality so we must take these cases.
We also know that #x!=0# and #x!=2# because then we would be dividing by #0#.
Case 1: Assume #x>0#. Then #x(x+2)>0#. Thus, no sign changes.
#2x^2-2x-2<=x^2+2x#
#x^2-4x-2<=0#
From this equation, we need to determine when it is less than 0. To do so, we can figure out where the quadratic intersects the x-axis. Since it opens up (the coefficient of the #x^2# term is greater than 0), we know that the x values between the zeroes are where the quadratic is negative.
#x=(-(-4)+-sqrt((-4)^2-4*1*(-2)))/(2*1)#
#x=(4+-sqrt(24))/2#
#x=2+-sqrt(6)#
That means the quadratic is negative for x-values on the interval #(2-sqrt(6)),(2+sqrt(6))#.
We only take the positive case because #2-sqrt(6)<0#, but we defined #x>0# for this case. Thus, we have found one interval:
#x:(0,2+sqrt(6)]#
Case 2; #x:(-2,0)#. If this is true, then #x<0# but #x+2>0#. Thus, the term #x(x+2)# is multiplying a negative and a positive, so #x(x+2)<0#. That means we have to flip the direction of the inequality when we multiply by #x(x+2)#.
#2x^2-2x-2>=x^2+2x#
#x^2-4x-2>=0#
Based on the math above, we know the quadratic is positive or zero when #x<=2-sqrt(6)# or #x>=2+sqrt(6)#. Since we know #x:(-2,0)#, we have found another interval.
#x:(-2,2-sqrt(6)]#.
Finally, case 3: #x<-2#. In this case, both #x# and #x+2# are negative, so their product is positive. That means our algebra and solutions will look very similar to Case 1.
#x:(2-sqrt(6)),(2+sqrt(6))#
Since the interval of solutions does not occur in our case of #x<-2#, there are no solutions in this case.
