Bài 3:
$a)$ `P = ((x\sqrt{x} -1)/(x -\sqrt{x}) -(x\sqrt{x} +1)/(x +\sqrt{x})) : ((2.(x -2\sqrt{x} +1))/(x -1)) ` `(ĐKXĐ: x ne 1; x ≥ 0)`
`= ((x\sqrt{x} -1).(x +\sqrt{x}) -(x\sqrt{x} +1).(x -\sqrt{x}))/((x -\sqrt{x}).(x +\sqrt{x})): ((2.(x -2\sqrt{x} +1))/(x -1))`
`= (x²\sqrt{x} +x² -x -\sqrt{x} -x²\sqrt{x} +x² -x +\sqrt{x})/((x -\sqrt{x}).(x +\sqrt{x})) : ((2.(\sqrt{x} -1)²)/(x -1))`
`= (2x² -2x)/((x -\sqrt{x}).(x +\sqrt{x})) . (x -1)/((2.(\sqrt{x} -1)²)`
`= (2x.(x -1).(x -1))/((x -\sqrt{x}).(x +\sqrt{x}).(2.(\sqrt{x} -1)²)`
`= (x.(x -1)^{2})/((x^{2} -x).(\sqrt{x} -1)²)`
`= (x.(x -1)^{2})/(x.(x -1).(\sqrt{x} -1)²`
`= (x -1)/((\sqrt{x} -1)²)`
`= ((\sqrt{x} -1).(\sqrt{x} +1))/((\sqrt{x} -1)²) = (\sqrt{x} +1)/(\sqrt{x} -1)`
$b)$
Để `P <0 ⇔ (\sqrt{x} +1)/(\sqrt{x} -1) < 0`
$⇔ \left[ \begin{array}{l}\begin{cases}\sqrt{x} +1 <0\\\sqrt{x} -1 >0\end{cases}\\\begin{cases} \sqrt{x} +1 >0\\\sqrt{x} -1 <0\end{cases}\end{array} \right.$
$⇔ \left[ \begin{array}{l}\begin{cases}\sqrt{x} <-1\\\sqrt{x} >1\end{cases}\\\begin{cases} \sqrt{x} >-1\\\sqrt{x} <1\end{cases}\end{array} \right.$
$⇔ \left[ \begin{array}{l}\begin{cases}x <-1\\x >1\end{cases} (Loại)\\\begin{cases} x >-1\\x <1\end{cases} (T/m)\end{array} \right.$
`⇒ -1 < x < 1 ⇒ x = 0`
$c) $
`P = (\sqrt{x} +1)/(\sqrt{x} -1) = 1 + 2/(\sqrt{x} -1)`
Để `P ∈ Z ⇔ 2/(\sqrt{x} -1) ∈ Z`
Mà $x ∈ Z$
`⇒ \sqrt{x} -1 ∈ Ư(2) = {±1; ±2}`
`⇒ \sqrt{x} ∈ {2; 0; 3; -1}`
Vì $\sqrt{x} ≥ 0$
`⇒ \sqrt{x} ∈ {2; 0; 3}`
`⇒ x ∈ {4; 0; 9}`
Vậy ......
