ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >4\end{matrix}\right.\)
\(M=A\cdot B=\dfrac{x}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)
=>\(M=\dfrac{x}{\sqrt{x}+2}\)
=>\(M=\dfrac{x-4+4}{\sqrt{x}+2}=\sqrt{x}-2+\dfrac{4}{\sqrt{x}+2}\)
=>\(M=\sqrt{x}+2+\dfrac{4}{\sqrt{x}+2}-4\)
=>\(M>=2\cdot\sqrt{\left(\sqrt{x}+2\right)\cdot\dfrac{4}{\sqrt{x}+2}}-4=0\)
Dấu '=' xảy ra khi \(\sqrt{x}+2=\sqrt{4}=2\)
=>\(\sqrt{x}=0\)
=>x=0(nhận)