It is a little unclear from the question exactly what is meant - the brackets are redundant. I'll supply answers for two possible question cases - a) Brackets removed; b) Brackets changed to modulus brackets.
a) Brackets removed
#1/x<=x-2#
Recall that multiplying through by a negative number implies changing the direction of an inequality, so:
If #x>=0#, then #1<=x(x-2)#, so #x^2-2x-1>=0#
If #x<0#, then #1>=x(x-2)#, so #x^2-2x-1<=0#
Solve the quadratic for #x^2-2x-1=0# by the quadratic formula:
#x=(2+-sqrt(4+4))/2=1+-sqrt(2)#
Then we know that the quadratic expression will change whether it satisfies the inequality or not at these points. Evaluate it at some sample points to check whether it is positive or negative on these intervals:
Let #f(x)=x^2-2x-1#. Then for the first interval (#x<1-sqrt(2)#), #f(-1)=1+2-1=2#, so #f(x)# is positive in this interval. For the second interval (#1-sqrt(2) < x < 1+sqrt(2)#), #f(0)=0-0-1=-1#, so #f(x)# is negative in this interval. For the third interval, #f(3)=9-6-1=2#, so #f(x)# is again positive.
Compare these to the piecewise pair of inequalities that we are solving:
The first interval (#x<1-sqrt(2)#) lies entirely within the negative #x# range, so we want #x^2-2x-1<=0#. But #f(x)# is positive on this interval, so there are no solutions.
The second interval (#1-sqrt(2) < x < 1+sqrt(2)#) spans #x=0#, so the needed inequality changes in the middle of the interval. #f(x)# is negative on this interval, so we fulfil our equation when we are at #x<0#, but not when #x>0#. Thus the range #1-sqrt(2) <= x <= 0# is one solution of the inequality.
The third interval (#1+sqrt(2) < x#) lies entirely within the positive #x# range, so we want #x^2-2x-1>=0#. We noted that #x# is positive on this interval, so #x>=1+sqrt(2)# is a second solution of the inequality.
In summary - two solution ranges for #x#:
#1-sqrt(2)<=x <0# and #x>=1+sqrt(2)#
Note that the expression is not valid at #x=0#, due to the presence of the #1/x# term.
b) Brackets changed to modulus brackets
#1/x<=|x-2|#
There are two cases here that we didn't have to worry about above:
If #x>=2#, then #1/x<=x-2#
If #x<2#, then #1/x<=-(x-2)#
Then we proceed as before, but combining these extra ranges into our working:
If #2<=x#, then #1<=x(x-2)#, so #x^2-2x-1>=0#
If #0<=x<2#, then #1<=-x(x-2)#, so #-x^2+2x-1>=0#
If #x<0#, then #1>=-x(x-2)#, so #-x^2+2x-1<=0#
We now have three ranges rather than the two we had before. One of them has the same quadratic expression as we had before; the other two have a similar expression with one coefficient sign change.
Solve the second quadratic for #-x^2+2x-1=0# by factorisation:
#-(x-1)^2=0rArrx=1#
This has two repeated roots at #x=1#. Consider the sign of the function #g(x)=-x^2+2x-1# - as it has only one root, it touches the #x# axis without passing through it - so it has the same sign over the whole function (excepting the single zero value) - evaluating it at e.g. 0 shows us quickly that that sign is negative.
Consider our three piecewise inequalities above.
When #x<0#, we need #-x^2+2x-1<=0#. We've established that the quadratic expression here is always negative or zero, so every #x# such that #x<0# is a solution.
When #0<= x <2#, we need #-x^2+2x-1>=0#. Only one point in this range fulfils this inequality: #x=1#, the zero point of #g(x)#.
When #2<=x#, we need #x^2-2x-1>=0#, the original quadratic expression, #f(x)#. Recall that the roots of #f(x)# are at #x=1+-sqrt(2)#. The higher of these is approximately 2.414, which is #>2#, so we will have a change in inequality satisfaction at this point too. From our answer above, we see that #f(x)>=0# when #x>=1+sqrt(2)#, and that it is negative between 2 and this point. So #x>=1+sqrt(2)# is a third answer to this problem.
In summary - three solution ranges to this problem, one of which is a single point:
#x<0#, #x=1#, and #x<=1+sqrt(2)#
Note as before that the inequality is undefined at #x=0# due to the presence of the #1/x# term.
Compare the solution cases between the two bracket possibilities. The solutions that give positive terms on both sides of the inequality are identical, as we would expect; otherwise they differ.
