Answer:
![\displaystyle -\frac{1}{2} \leq x < 1](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%20-%5Cfrac%7B1%7D%7B2%7D%20%5Cleq%20x%20%3C%201)
Step-by-step explanation:
<u>Inequalities</u>
They relate one or more variables with comparison operators other than the equality.
We must find the set of values for x that make the expression stand
![\displaystyle \frac{6x^5+11x^4-11x-6}{(2x^2-3x+1)^2} \leq 0](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B6x%5E5%2B11x%5E4-11x-6%7D%7B%282x%5E2-3x%2B1%29%5E2%7D%20%5Cleq%200)
The roots of numerator can be found by trial and error. The only real roots are x=1 and x=-1/2.
The roots of the denominator are easy to find since it's a second-degree polynomial: x=1, x=1/2. Hence, the given expression can be factored as
![\displaystyle \frac{(x-1)(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)^2(x-\frac{1}{2})^2} \leq 0](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B%28x-1%29%28x%2B%5Cfrac%7B1%7D%7B2%7D%29%286x%5E3%2B14x%5E2%2B10x%2B12%29%7D%7B%28x-1%29%5E2%28x-%5Cfrac%7B1%7D%7B2%7D%29%5E2%7D%20%5Cleq%200)
Simplifying by x-1 and taking x=1 out of the possible solutions:
![\displaystyle \frac{(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)(x-\frac{1}{2})^2} \leq 0](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B%28x%2B%5Cfrac%7B1%7D%7B2%7D%29%286x%5E3%2B14x%5E2%2B10x%2B12%29%7D%7B%28x-1%29%28x-%5Cfrac%7B1%7D%7B2%7D%29%5E2%7D%20%5Cleq%200)
We need to find the values of x that make the expression less or equal to 0, i.e. negative or zero. The expressions
![(6x^3+14x^2+10x+12)](https://tex.z-dn.net/?f=%286x%5E3%2B14x%5E2%2B10x%2B12%29)
is always positive and doesn't affect the result. It can be neglected. The expression
![(x-\frac{1}{2})^2](https://tex.z-dn.net/?f=%28x-%5Cfrac%7B1%7D%7B2%7D%29%5E2)
can be 0 or positive. We exclude the value x=1/2 from the solution and neglect the expression as being always positive. This leads to analyze the remaining expression
![\displaystyle \frac{(x+\frac{1}{2})}{(x-1)} \leq 0](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B%28x%2B%5Cfrac%7B1%7D%7B2%7D%29%7D%7B%28x-1%29%7D%20%5Cleq%200)
For the expression to be negative, both signs must be opposite, that is
![(x+\frac{1}{2})\geq 0, (x-1)](https://tex.z-dn.net/?f=%28x%2B%5Cfrac%7B1%7D%7B2%7D%29%5Cgeq%200%2C%20%28x-1%29%3C0)
Or
![(x+\frac{1}{2})\leq 0, (x-1)>0](https://tex.z-dn.net/?f=%28x%2B%5Cfrac%7B1%7D%7B2%7D%29%5Cleq%200%2C%20%28x-1%29%3E0)
Note we have excluded x=1 from the solution.
The first inequality gives us the solution
![\displaystyle -\frac{1}{2} \leq x < 1](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%20-%5Cfrac%7B1%7D%7B2%7D%20%5Cleq%20x%20%3C%201)
The second inequality gives no solution because it's impossible to comply with both conditions.
Thus, the solution for the given inequality is
![\boxed{\displaystyle -\frac{1}{2} \leq x < 1 }](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cdisplaystyle%20%20-%5Cfrac%7B1%7D%7B2%7D%20%5Cleq%20x%20%3C%201%20%7D)