a)Thay \(x=36\) vào A ta được:
\(B=\dfrac{\sqrt{x}}{\sqrt{x}-3}=\dfrac{\sqrt{36}}{\sqrt{36}-3}=2\)
b)Với \(x>0;x\ne1\)
\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(A=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(A=\dfrac{\sqrt{x}+2}{\sqrt{x}}\)
c)\(P=A\cdot B=\dfrac{\sqrt{x}+2}{\sqrt{x}}\cdot\dfrac{\sqrt{x}}{\sqrt{x}-3}=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\) với \(x>0;x\ne9\)
\(P=\dfrac{\sqrt{x}-3+5}{\sqrt{x}-3}=1+\dfrac{5}{\sqrt{x}-3}\)
Để \(P\) nguyên \(\Rightarrow\) \(\dfrac{5}{\sqrt{x}-3}\in Z\)
\(\Rightarrow\dfrac{5}{\sqrt{x}-3}\inƯ\left(5\right)=\left\{-1;1;-5;5\right\}\)
\(\Rightarrow x=\left\{64;4\right\}\)
