1.) x3+27/x2-3x-10 < 0
2.) (3x-1) (2-5x)/2x+1 > 0
3.) 2/x+2< or equal to x/x+1
I'm assuming this is
1.) (x^3+27) / (x^2-3x-10) < 0 factoring the numerator and denominator, we have
[(x + 3) (x^2 - 3x + 9)] / [ ( x - 5) (x + 2) ] < 0
The second factor in the numerator is always positive so the numerator is < 0 whenever x < -3
The denominator is undefined whenever x = 5 or x = -2
And the denominator will < 0 whenever -2 < x < 5
So......the function will be < 0 whenever the numerator is positive and the denominator is negative or when the numerator is negative and the denominator is positive
The first situation will occur when -2 < x < 5
And the second wil occur when x < - 3
So the solution set is [ x < - 3 ] U [ -2 < x < 5 ]
Here's the graph :
https://www.desmos.com/calculator/xgh0b1qmvm
I'm assuming that we have :
2.) (3x-1) (2-5x) / (2x+1) > 0
Setting each factor to 0 and solving for x, the critical values are :
x = 1/3 , x = 2/5 and x = -1/2
The function will be undefined when x = - 1 /2
The numerator is positive when 1/3 < x < 2/5
And the denominator is positive when x > -1/2
And the function is > 0 whenever the numerator and denominator have the same sign
So...the function is > 0 when
1/3 < x < 2/5 or when x < -1/2
Here's the graph :
https://www.desmos.com/calculator/utup8jcqp2
I'm assuming that the last one is :
2 / (x+2) ≤ x / (x+1)
Setting each factor to 0 and solving for x, the critical values are -2 and - 1
Note that the inequality is undefined at these two x values
Turn this into an equality
Cross-multiplying, we have
2 (x + 1) = x ( x + 2) simplify
2x + 2 = x^2 + 2x subtract 2x from both sides
2 = x^2 rearranging, we have
x^2 - 2 = 0
And this will be true when x = - √2 or x = √2
And these are the other two critical values that we need
Note that the inequality will be true when
x < -2 , - √2 ≤ x < -1 and x ≥ √2
Here's a graph :
https://www.desmos.com/calculator/zu0vmqfgux