Giải thích các bước giải:
a.Ta có:
$R=1:\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right)$
$\to R=1:\left(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)$
$\to R=1:\left(\dfrac{x+2+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)$
$\to R=1:\left(\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)$
$\to R=1:\left(\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)$
$\to R=1:\left(\dfrac{\left(\sqrt{x}-1\right)\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)$
$\to R=1:\dfrac{\sqrt{x}}{x+\sqrt{x}+1}$
$\to R=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}$
$\to R=\sqrt{x}+1+\dfrac{1}{\sqrt{x}}$
b.Ta có:
$R-3=\sqrt{x}+1+\dfrac{1}{\sqrt{x}}-3$
$\to R-3=\sqrt{x}-2+\dfrac{1}{\sqrt{x}}$
$\to R-3=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}}$
$\to R-3=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\ge 0$
$\to R\ge 3$
c.Để $R>4$
$\to R-4>0$
$\to \sqrt{x}+1+\dfrac{1}{\sqrt{x}}-4>0$
$\to \sqrt{x}-3+\dfrac{1}{\sqrt{x}}>0$
$\to x-3\sqrt{x}+1>0$
$\to \left(x-2\sqrt{x}\cdot\dfrac32+\dfrac94\right)>\dfrac54$
$\to \left(\sqrt{x}-\dfrac32\right)^2>\dfrac54$
$\to \sqrt{x}-\dfrac32>\dfrac{\sqrt{5}}{2}$
$\to \sqrt{x}>\dfrac{3+\sqrt{5}}{2}$
$\to x>\left(\dfrac{3+\sqrt{5}}{2}\right)^2$
Hoặc $\sqrt{x}-\dfrac32<-\dfrac{\sqrt{5}}{2}$
$\to x<\left(\dfrac{-\sqrt{5}+3}{2}\right)^2$
$\to 0<x<\left(\dfrac{-\sqrt{5}+3}{2}\right)^2$
