1) Với x=4 thì
\(A=\dfrac{2\sqrt{4}}{\sqrt{4}+3}=\dfrac{4}{2+3}=\dfrac{4}{5}\)
2) \(P=A+B\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{x-9}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x+3}\right)}\)
\(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)
3) Để P< 3 thì
\(\dfrac{3\sqrt{x}}{\sqrt{x}-3}< 3\)
\(\Leftrightarrow\dfrac{3\sqrt{x}}{\sqrt{x}-3}-\dfrac{3\left(\sqrt{x}-3\right)}{\sqrt{x}-3}< 0\)
\(\Leftrightarrow\dfrac{9}{\sqrt{x}-3}< 0\)
\(\Rightarrow\sqrt{x}-3< 0\) ( vì 9>0)
<=> x<9
Vậy giá trị nguyên lớn nhất của x để P <3 là 8