1) Thay x=9 vào A ta có:
\(A=\dfrac{\sqrt{9}+1}{\sqrt{9}-2}=\dfrac{3+1}{3-2}=\dfrac{4}{1}=4\)
2) \(B=\dfrac{2x+8\sqrt{x}}{x-4}-\dfrac{2}{\sqrt{x}+2}-\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\)
\(B=\dfrac{2x+8\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2}{\sqrt{x}+2}-\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\)
\(B=\dfrac{2x+8\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(B=\dfrac{2x+8\sqrt{x}-2\sqrt{x}+4-x+2\sqrt{x}+8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{x+8\sqrt{x}+12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{x+6\sqrt{x}+2\sqrt{x}+12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{\sqrt{x}\left(\sqrt{x}+6\right)+2\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{\sqrt{x}+6}{\sqrt{x}-2}\)
3) Ta có: \(P=\dfrac{B}{A}\)
\(P=\dfrac{\sqrt{x}+6}{\sqrt{x}-2}:\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)
\(P=\dfrac{\sqrt{x}+6}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
\(P=\dfrac{\sqrt{x}+6}{\sqrt{x}+1}\)
P nguyên khi:
\(\dfrac{\sqrt{x}+6}{\sqrt{x}+1}\) nguyên
Mà: \(\dfrac{\sqrt{x}+6}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1+5}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{\sqrt{x}+1}+\dfrac{5}{\sqrt{x}+1}=1+\dfrac{5}{\sqrt{x}+1}\)
\(\Rightarrow5\) ⋮ \(\sqrt{x}+1\)
\(\sqrt{x}+1\inƯ\left(5\right)=\left\{5;-5;1;-1\right\}\)
\(x\ge0\) nên: \(\sqrt{x}+1\in\left\{5;1\right\}\)
\(\Rightarrow x\in\left\{16;0\right\}\)
