a: \(D=\left(\dfrac{\left(2+\sqrt{x}\right)^2-\left(2-\sqrt{x}\right)^2+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right)\cdot\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\)
\(=\dfrac{4+4\sqrt{x}+x-4+4\sqrt{x}-x+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\cdot\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\)
\(=\dfrac{4x+8\sqrt{x}}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+3}=\dfrac{4x}{\sqrt{x}+3}\)
b: ta có: |D|=1
=>D=1(nhận) hoặc D=-1(loại) (Vì \(D=\dfrac{4x}{\sqrt{x}+3}>0\) với mọi x thỏa mãn ĐKXĐ)
\(\Leftrightarrow4x=\sqrt{x}+3\)
\(\Leftrightarrow4x-\sqrt{x}-3=0\)
\(\Leftrightarrow4x-4\sqrt{x}+3\sqrt{x}-3=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(4\sqrt{x}+3\right)=0\)
=>x=1(nhận)