ĐKXĐ: \(x\ge0;x\ne4\)
\(P=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\frac{x+5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-3\sqrt{x}+2-\left(x+4\sqrt{x}+3\right)-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\frac{-x-7\sqrt{x}-6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{-\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+6}{2-\sqrt{x}}\)
b/ \(P=\frac{\sqrt{x}-2+8}{2-\sqrt{x}}=-1+\frac{8}{2-\sqrt{x}}\in Z\)
\(\Rightarrow2-\sqrt{x}=Ư\left(8\right)=\left\{-8;-4;-2;-1;1;2\right\}\) (do \(2-\sqrt{x}\le2\) nên chỉ cần xét các ước trên)
\(\Rightarrow\sqrt{x}=\left\{10;6;4;3;1;0\right\}\)
\(\Rightarrow x=\left\{0;1;9;16;36;100\right\}\)
c/ \(\frac{\sqrt{x}+6}{2-\sqrt{x}}\ge1\Leftrightarrow\frac{\sqrt{x}+6}{2-\sqrt{x}}-1\ge0\Leftrightarrow\frac{2\sqrt{x}+4}{2-\sqrt{x}}\ge0\)
\(\Leftrightarrow2-\sqrt{x}>0\Leftrightarrow0\le x< 4\)
