Đkxđ: $\left \{ {{\frac{x+1}{x}≥0} \atop {x\neq0;x\neq-1}} \right.$
⇔\(\left[ \begin{array}{l}x>0\\x<-1\end{array} \right.\)
Đặt: $\sqrt{\frac{x+1}{x}}$ =t (t>0)
Bpt ⇔ $\frac{1}{t^2}$ -2t >3$
$⇔1-2t^3>3t^2$$⇔2t^3+3t^2-1<0$
$⇔2t^3+4t^2+2t-t^2-2t-1<0$
$⇔2t(t+1)^2-(t+1)^2<0$
$⇔(2t-1)(t+1)^2<0$
$⇔2t-1<0$$⇔t<\frac{1}{2}$
Suy ra:$ \frac{x+1}{x}<\frac{1}{4} ⇔\frac{x+1}{x}-\frac{1}{4}<0$
$⇔\frac{3x+4}{4x}<0$
$⇔\frac{-4}{3}<x<0$
Kết hợp vs đkxđ: ⇒ $⇔\frac{-4}{3}<x<-1$
