ĐK \(x\ge0\)

Ta có \(\frac{x-\sqrt{3}x+3}{x\sqrt{x}+3\sqrt{3}}=\frac{x-\sqrt{3}x+3}{\left(\sqrt{x}+\sqrt{3}\right)\left(x-\sqrt{3}x+3\right)}=\frac{1}{\sqrt{x}+\sqrt{3}}\)

ĐK \(\hept{\begin{cases}x^2-x\ge0\\3-x\ge0\end{cases}\Rightarrow\orbr{\begin{cases}x\le0\\1\le x\le3\end{cases}}}\)

Phương trình \(\Rightarrow x^2-x=3-x\Rightarrow x^2=3\Rightarrow\orbr{\begin{cases}x=\sqrt{3}\\x=-\sqrt{3}\end{cases}\left(tm\right)}\)

Vậy \(\orbr{\begin{cases}x=\sqrt{3}\\x=-\sqrt{3}\end{cases}}\)

\(x^2-x-1000\sqrt{8000x+1}=1000\)

\(\Leftrightarrow\left(x^2-x-4002000\right)-\left(1000\sqrt{8000x+1}-4001000\right)=0\)

\(\Leftrightarrow\left(x-2001\right)\left(x+2000\right)-\frac{1000\left(8000x+1-4001^2\right)}{1000\sqrt{8000x+1}+4001000}=0\)

\(\Leftrightarrow\left(x-2001\right)\left(x+2000\right)-\frac{1000\cdot8000\left(x-2001\right)}{1000\sqrt{8000x+1}+4001000}=0\)

\(\Leftrightarrow\left(x-2001\right)\left(\left(x+2000\right)-\frac{1000\cdot8000}{1000\sqrt{8000x+1}+4001000}\right)=0\)

\(\Rightarrow x=2001\)

\(\)còn cái trong ngoặc thì sao

\(=\frac{\sqrt{3}+\sqrt{2}-1}{\left(\sqrt{3}+\sqrt{2}+1\right)\left(\sqrt{3}+\sqrt{2}-1\right)}=\frac{\sqrt{3}+\sqrt{2}-1}{\left(\sqrt{3}+\sqrt{2}\right)^2-1}=\)

\(=\frac{\sqrt{3}+\sqrt{2}-1}{3+2\sqrt{3}\sqrt{2}+2-1}=\frac{\sqrt{3}+\sqrt{2}-1}{2\sqrt{3}\sqrt{2}+4}=\frac{\left(\sqrt{3}+\sqrt{2}-1\right)\left(2\sqrt{3}\sqrt{2}-4\right)}{4.3.2-16}\)

\(=\frac{2.3.\sqrt{2}-4.\sqrt{3}+2.2.\sqrt{3}-4.\sqrt{2}-2.\sqrt{3}.\sqrt{2}+4}{8}=\)

\(=\frac{2.\sqrt{2}-2.\sqrt{3}.\sqrt{2}+4}{8}=\frac{\sqrt{2}-\sqrt{6}+2}{4}\)