Answer :

UnicornAyushh

Answer:

[tex]\implies\sf{(\frac{2x}{2x^2 + 5x + 2} > \frac{1}{x+1})}[/tex]

[tex]\implies\sf{(\frac{2x}{2x^2 + 4x + x + 2} > \frac{1}{x+1})}[/tex]

[tex]\implies\sf{(\frac{2x}{(2x+1)(x+2)} - \frac{1}{x+1} > 0)}[/tex]

[tex]\implies\sf{(\frac{2x(x+1) - 2x^2 - 5x - 2}{(2x+1)(x+2)(x+1)} > 0)}[/tex]

[tex]\implies\sf{(\frac{- (3+2)}{(2x+1)(x+2)(x+1)} > 0)}[/tex]

[tex]\implies\sf{x \in (-2,-1) \cup (-\frac{2}{3}, -\frac{1}{2})}[/tex]

[tex]\implies\boxed{\sf{-2<x<-1}}[/tex]

mathdude500

Answer:

[tex]\boxed{\bf\: - 2 < x < - 1 \: } \\ [/tex]

Step-by-step explanation:

Given inequality is

[tex]\sf\: \dfrac{2x}{ {2x}^{2} + 5x + 2} > \dfrac{1}{x + 1} \\ [/tex]

[tex]\sf\: \dfrac{2x}{ {2x}^{2} + 4x + x + 2} > \dfrac{1}{x + 1} \\ [/tex]

[tex]\sf\: \dfrac{2x}{ 2x(x + 2) + (x + 2)} > \dfrac{1}{x + 1} \\ [/tex]

[tex]\sf\: \dfrac{2x}{ (x + 2) (2x + 1)} > \dfrac{1}{x + 1} \: \: \left\{x \ne \: - 1, - 2, - \dfrac{1}{2} \right\} \\ [/tex]

[tex]\sf\: \dfrac{2x}{ (x + 2) (2x + 1)} - \dfrac{1}{x + 1} > 0 \: \\ [/tex]

[tex]\sf\: \dfrac{2x(x + 1) - (x + 2)(2x + 2)}{ (x + 2) (2x + 1)(x + 1)} > 0 \: \\ [/tex]

[tex]\sf\: \dfrac{ {2x}^{2} + 2x - {2x}^{2} - 5x - 2 }{ (x + 2) (2x + 1)(x + 1)} > 0 \: \\ [/tex]

[tex]\sf\: \dfrac{ - 3x - 2 }{ (x + 2) (2x + 1)(x + 1)} > 0 \: \\ [/tex]

[tex]\sf\: \dfrac{ - (3x + 2 )}{ (x + 2) (2x + 1)(x + 1)} > 0 \: \\ [/tex]

[tex]\sf\: \dfrac{3x + 2 }{ (x + 2) (2x + 1)(x + 1)} < 0 \: \\ [/tex]

[tex]\implies\sf\:x \: \in \: ( - 2, - 1) \: \cup \: \bigg( - \dfrac{2}{3}, - \dfrac{1}{2} \bigg) \\ [/tex]

Hence,

[tex] \implies\bf\: - 2 < x < - 1 \: satisfies \: \dfrac{2x}{ {2x}^{2} + 5x + 2} > \dfrac{1}{x + 1} \\ [/tex]

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