What is (2x+1)/x-x/(x+2)<= (x^2+2x+4)/(x^2+2x) ?

The solution to (2x+1)/x-x/(x+2)<= (x^2+2x+4)/(x^2+2x) is x<-2\lor 0<x<= 2/3

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(2x+1)/x-x/(x+2)<= (x^2+2x+4)/(x^2+2x)

\frac{2 x + 1}{x} - \frac{x}{x + 2} \leq \frac{x ^{2} + 2 x + 4}{x ^{2} + 2 x}

x < - 2 or 0 < x \leq \frac{2}{3}

Interval Notation

(- \infty, - 2) \cup (0, \frac{2}{3}]

Decimal Notation

x < - 2 or 0 < x \leq 0.66666 \dots

Solution steps

\frac{2 x + 1}{x} - \frac{x}{x + 2} \leq \frac{x ^{2} + 2 x + 4}{x ^{2} + 2 x}

Rewrite in standard form

\frac{3 x - 2}{x ( x + 2 )} \leq 0

Identify the intervals

x < - 2 or 0 < x \leq \frac{2}{3}

\frac{1}{3 + 4 i}

\frac{6 i}{9 - 9 i}

(2 x + 1)^{2} = 12

4 v^{2} - 64

(\frac{3 x}{y})^{- 3}

What is (2x+1)/x-x/(x+2)<= (x^2+2x+4)/(x^2+2x) ?
The solution to (2x+1)/x-x/(x+2)<= (x^2+2x+4)/(x^2+2x) is x<-2\lor 0<x<= 2/3