a: ĐKXĐ: x>0; x<>9
b: \(M=\left(\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{3\sqrt{x}+1-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{3\sqrt{x}+1-\sqrt{x}+3}\)
\(=\frac{-3\sqrt{x}-9}{\sqrt{x}+3}\cdot\frac{\sqrt{x}}{2\sqrt{x}+4}=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
c: 2M là số nguyên
=>\(2\cdot\left(-3\right)\sqrt{x}\) ⋮\(2\sqrt{x}+4\)
=>\(-6\sqrt{x}\) ⋮\(2\sqrt{x}+4\)
=>\(-6\sqrt{x}-12+12\) ⋮\(2\sqrt{x}+4\)
=>12⋮\(2\sqrt{x}+4\)
=>\(2\sqrt{x}+4\in\left\lbrace4;6;12\right\rbrace\)
=>\(\sqrt{x}+2\) ∈{2;3;6}
=>\(\sqrt{x}\) ∈{0;1;4}
=>x∈{0;1;16}
Kết hợp ĐKXĐ, ta được: x∈{1;16}
d: M<=-1
=>M+1<=0
=>\(\frac{-3\sqrt{x}+2\sqrt{x}+4}{2\sqrt{x}+4}\le0\)
=>\(\frac{-\sqrt{x}+4}{2\sqrt{x}+4}\le0\)
=>\(-\sqrt{x}+4\le0\)
=>\(-\sqrt{x}\le-4\)
=>\(\sqrt{x}\ge4\)
=>x>=16
