a/ \(B=\frac{\sqrt{x}+3}{\sqrt{x}+1}-\frac{5}{1-\sqrt{x}}+\frac{4}{x-1}=\frac{\sqrt{x}+3}{\sqrt{x}+1}+\frac{5}{\sqrt{x}-1}+\frac{4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)+5\left(\sqrt{x}+1\right)+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{x+2\sqrt{x}-3+5\sqrt{x}+5+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x+7\sqrt{x}+6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+6}{\sqrt{x}-1}\)
b/ \(B< 1\Leftrightarrow\frac{\sqrt{x}+6}{\sqrt{x}-1}< 1\Leftrightarrow\frac{\sqrt{x}+6}{\sqrt{x}-1}-1< 0\)
\(\Leftrightarrow\frac{\sqrt{x}+6-\sqrt{x}+1}{\sqrt{x}-1}< 0\Leftrightarrow\frac{7}{\sqrt{x}-1}< 0\)
\(\Leftrightarrow\sqrt{x}-1< 0\left(vì7>0\right)\)
\(\Leftrightarrow x< 1\)
c/ Ta có: \(A.B=\frac{\sqrt{x}-1}{\sqrt{x}+1}.\frac{\sqrt{x}+6}{\sqrt{x}-1}=\frac{\sqrt{x}+6}{\sqrt{x}+1}=\frac{\sqrt{x}+1+5}{\sqrt{x}+1}=1+\frac{5}{\sqrt{x}+1}\)
Để A.B nhận giá trị nguyên thì: \(5⋮\left(\sqrt{x}+1\right)\Leftrightarrow\left(\sqrt{x}+1\right)\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Giải ra tìm được \(x=\left\{0;2\right\}\)