Đkxđ : \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
a) \(P=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{6\sqrt{x}-4}{x-1}+\frac{3}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x+\sqrt{x}-6\sqrt{x}+4+3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
b) Ta có :
\(P=\frac{\sqrt{x}-1}{\sqrt{x}+1}=\frac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\frac{2}{\sqrt{x}+1}\)
Để P nhận giá trị nguyên thì :
\(\sqrt{x}+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
⇔ x ϵ { 0 ; 4 ; 9 }
