[tex] \sqrt{ \frac{2 - 3x}{x + 4} } > - 2[/tex]

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вопрос

To solve this inequality, we first note that the square root of a fraction is only defined when the numerator is non-negative and the denominator is positive. Therefore, we need to consider two cases:

Case 1: ( x + 4 > 0 ) and ( 2 - 3x \geq 0 )
( x + 4 > 0 ) implies ( x > -4 )
( 2 - 3x \geq 0 ) implies ( 2 \geq 3x ) and ( x \leq \frac{2}{3} )

Therefore, the solution to this case is: ( -4 < x \leq \frac{2}{3} )

Case 2: ( x + 4 < 0 ) and ( 2 - 3x \leq 0 )
( x + 4 < 0 ) implies ( x < -4 )
( 2 - 3x \leq 0 ) implies ( 2 \leq 3x ) and ( x \geq \frac{2}{3} )

However, in this case, the square root of the fraction is always non-negative, so it will always be greater than -2.

Therefore, the overall solution to the inequality is: ( -4 < x \leq \frac{2}{3} )