1) Thay x=9 vào B ta có:
\(A=\dfrac{5\cdot\sqrt{9}}{\sqrt{9}+2}=\dfrac{5\cdot3}{5+2}=\dfrac{15}{7}\)
2) \(B=\dfrac{5}{\sqrt{x}-2}+\dfrac{16+2\sqrt{x}}{4-x}\)
\(B=\dfrac{5}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+16}{x-4}\)
\(B=\dfrac{5\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{2\sqrt{x}+16}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{5\sqrt{x}+10-2\sqrt{x}-16}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{3\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{3}{\sqrt{x}+2}\)
3) Mà: \(A+B=\dfrac{3}{\sqrt{x}+2}+\dfrac{5\sqrt{x}}{\sqrt{x}+2}=\dfrac{5\sqrt{x}+3}{\sqrt{x}+2}\)
\(\Rightarrow\dfrac{5\sqrt{x}+3}{\sqrt{x}+2}< 3\)
\(\Rightarrow\dfrac{5\sqrt{x}+3-3\left(\sqrt{x}+2\right)}{\sqrt{x}+2}< 0\)
\(\Rightarrow\dfrac{2\sqrt{x}-3}{\sqrt{x}+2}< 0\)
\(\Rightarrow2\sqrt{x}< 3\)
\(\Rightarrow x< \dfrac{9}{4}\)
\(\Rightarrow0\le x< \dfrac{9}{4}\)
Nên x nguyên lớn nhất là: \(x=2\)
