ĐKXĐ: \(x>0;x\ne1\)
\(A=\frac{x\sqrt{x}-2x}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{2x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x\sqrt{x}-2x+x-1+2x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{x\sqrt{x}+x-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(x+\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\)
Đặt \(\sqrt{x}=a>0\) ; \(a\ne1\)
\(\Rightarrow A=\frac{a+2}{a^2+a+1}\Rightarrow A.a^2+\left(A-1\right)a+A-2=0\)
\(\Delta=\left(A-1\right)^2-4A\left(A-2\right)\ge0\)
\(\Leftrightarrow-3A^2+6A+1\ge0\)
\(\Rightarrow\frac{3-2\sqrt{3}}{3}\le A\le\frac{3+2\sqrt{3}}{3}\)
Hơn nữa do A nguyên và A hiển nhiên dương nên \(A=\left\{1;2\right\}\)
- Với \(A=1\Rightarrow\frac{a+2}{a^2+a+1}=1\Rightarrow a^2=1\Rightarrow a=1\Rightarrow x=1\)
- Với \(A=2\Rightarrow\frac{a+2}{a^2+a+1}=2\Rightarrow2a^2+a=0\) (vô nghiệm)
