Tks vì cho điểm nha
$ĐKXĐ: x \geq 0$
$P=\dfrac{-\sqrt[]x-1+1}{\sqrt[]x+1}$
$=\dfrac{1}{\sqrt[]x+1}-1$
Nên $P$ nguyên $⇔\dfrac{1}{\sqrt[]x+1}$ nguyên
Có: $\sqrt[]{x}≥0∀x∈ĐKXĐ$
Nên $\sqrt[]x+1≥1∀x∈ĐKXĐ$
Suy ra $\dfrac{1}{\sqrt[]x+1}≤1$
Mà $\dfrac{1}{\sqrt[]x+1}>0$ (do $\sqrt[]x+1>0$)
$→0<\dfrac{1}{\sqrt[]x+1}≤1$
$\dfrac{1}{\sqrt[]x+1}$ nguyên $⇒\dfrac{1}{\sqrt[]x+1}=1$
$⇔\sqrt[]x+1=1$
$⇔\sqrt[]x=0$
$⇔x=0$
Vậy $x=0$ t/m đề
