Question

Solve this inequality :-
$\sqrt{ \frac{ {x}^{2} + x + 1 }{ {x}^{2} - x - 2} } > 1$
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Answer 1

Given:

$\sqrt{\dfrac{x^2 + x + 1}{x^2 - x - 2} } > 1$

To Find:

The value of x

Solution:

$\sqrt{\dfrac{x^2 + x + 1}{x^2 - x - 2} } > 1$

Square both sides:

$\dfrac{x^2 + x + 1}{x^2 - x - 2} > 1^2$

$\dfrac{x^2 + x + 1}{x^2 - x - 2} > 1$

Cross multiply:

$x^2 + x + 1 >x^2 - x -2$

$2x > -3$

$x> - \dfrac{3}{2}$

Answer: x > -3/2

Answer 2

Step-by-step explanation:

$\huge\mathfrak{Given}$

$\sqrt{ \frac{ {x}^{2} + x + 1}{ {x}^{2} - x - 2} } > 1$

$\huge\mathfrak{solution}$

$\sqrt{ \frac{ {x}^{2} + x + 1}{ {x}^{2} - x - 2} } > 1$

square both side

$\frac{ {x}^{2} + x + 1}{ {x}^{2} - x - 2} > {1}^{2}$

$\frac{ {x}^{2} + x + 1 }{ {x}^{2} - x - 2 } > 1$

cross multiply

${x}^{2} + x + 1 > {x}^{2} - x - 2 \\ 2x > - 3 \\ x > - \frac{ 3}{2}$

Answer

$x > - \frac{3}{2}$

Hope it helpful to you

mark brainliest