\(c,P=\dfrac{1}{3}\\ \Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{1}{3}\\ \Leftrightarrow3\sqrt{x}-3=\sqrt{x}+1\\ \Leftrightarrow2\sqrt{x}=4\\ \Leftrightarrow\sqrt{x}=2\\ \Leftrightarrow x=4\)
\(d,\) Ta có:
\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)
\(P\) nguyên \(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\) nguyên
\(\Leftrightarrow\dfrac{2}{\sqrt{x}+1}\) nguyên \(\Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{\pm2;\pm1\right\}\)
Ta có bảng sau:
| \(\sqrt{x}+1\) | \(-2\) | \(-1\) | \(1\) | \(2\) |
| \(\sqrt{x}\) | \(-3\) | \(-2\) | \(0\) | \(1\) |
| \(x\) | Loại | Loại | \(0\) (Loại) | \(1\) |
Vậy \(P\) nguyên \(\Leftrightarrow x=1\)
a là 0 hoặc 1