\(M=\frac{\sqrt{x}+2+\sqrt{x}-2-x}{4-x}=\frac{2\sqrt{x}-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{-\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{-\sqrt{x}}{\sqrt{x}+2}\)
ĐKXĐ: \(x\ge0;x\ne4\)
c/ thay x=\(6-4\sqrt{2}\) vào M có:
\(\frac{-\sqrt{6-4\sqrt{2}}}{\sqrt{6-4\sqrt{2}}+2}=\frac{-\sqrt{\left(2+\sqrt{2}\right)^2}}{\sqrt{\left(2+\sqrt{2}\right)^2}+2}=\frac{-2-\sqrt{2}}{4+\sqrt{2}}=\frac{-3-\sqrt{2}}{7}\)
d/ Để M=\(\frac{3}{2}\)\(\Leftrightarrow\frac{-\sqrt{x}}{\sqrt{x}+2}=\frac{3}{2}\Leftrightarrow-2\sqrt{x}=3\sqrt{x}+6\Leftrightarrow5\sqrt{x}=-6\left(vl\right)\)
Vậy ko tồn tai x để M=3/2
