# `(2x +1 - (21x +39)) / (х ^ 2 + х-2)> = 1 / (х +2)`

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### 2 Answers

You need to bring the terms to a common denominator, such that:

`((2x + 1 - (2x + 39))(x + 2))/((x+2)(x^2 + x + 2)) - (x^2 + x + 2)/((x+2)(x^2 + x + 2)) >= 0`

`(-38x - 76 - x^2 - x - 2)/((x+2)(x^2 + x + 2)) >= 0`

`(-x^2 - 39x - 78)/((x+2)(x^2 + x + 2)) >= 0`

You should multiplicate by -1 changing the direction of inequality, such that:

`(x^2 + 39x + 78)/((x+2)(x^2 + x + 2)) <= 0`

The fraction `(x^2 + 39x + 78)/((x+2)(x^2 + x + 2))` is negative if `x^2 + 39x + 78 < 0` and `(x+2)(x^2 + x + 2) > 0` or `x^2 + 39x + 78 > 0` and `(x+2)(x^2 + x + 2) < 0` .

You need to consider the first system of inequalities, such that:

`{(x^2 + 39x + 78 < 0),((x+2)(x^2 + x + 2) > 0):}`

You need to attach the equations `{(x^2 + 39x + 78 = 0),((x+2)(x^2 + x + 2) = 0):}`

You need to solve the top equation using quadratic formula,such that:

`x_(1,2) = (-39+-sqrt(1521 - 312))/2 => x_(1,2) = (-39+-sqrt(1209))/2`

The inequality stays negative for `x in ((-39-sqrt(1209))/2,(-39+sqrt(1209))/2)` .

You need to solve the inequality `(x+2)(x^2 + x + 2) > 0` , hence, since `x^2 + x + 2 > 0` , then `x + 2 > 0 => x > -2.`

The solution to the system of inequalities `{(x^2 + 39x + 78 < 0),((x+2)(x^2 + x + 2) > 0):}` is `x in (-2, oo).`

You need to consider the first system of inequalities, such that:

`{(x^2 + 39x + 78 > 0),((x+2)(x^2 + x + 2) < 0):}`

`x in (-oo, (-39-sqrt(1209))/2)U((-39+sqrt(1209))/2,oo) nn (-oo,-2) => x in (-oo, (-39-sqrt(1209))/2).`

**Hence, evaluating the solutions to the given inequality yields `x in (-2, oo)` or **`x in (-oo, (-39-sqrt(1209))/2).`

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К сожалению, фатальная ошибка) 2x +1- (21x +39) / (х ^ 2 + х-2)> = -1 / (х +2)

sorry, fatal error) 2x+1-(21x+39)/(x^2+x-2)>=-1/(x+2)