`1)`\(x=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\)
Thế vào `A` ta được:
\(A=\dfrac{\sqrt{\left(\sqrt{5}+1\right)^2}+3}{\sqrt{\left(\sqrt{5}+1\right)^2}-2}=\dfrac{\left|\sqrt{5}+1\right|+3}{\left|\sqrt{5}+1\right|-2}=\dfrac{\sqrt{5}+1+3}{\sqrt{5}+1-2}=\dfrac{\sqrt{5}+4}{\sqrt{5}-1}\)
`2)`\(B=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+2}-\dfrac{4x}{4-x}\)
\(B=\dfrac{\left(\sqrt{x}+2\right)^2-\left(\sqrt{x}-2\right)^2+4x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(B=\dfrac{x+4\sqrt{x}+4-x+4\sqrt{x}-4+4x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(B=\dfrac{4x+8\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(B=\dfrac{4\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(B=\dfrac{4\sqrt{x}}{\sqrt{x}-2}\)
`3)`\(P=\dfrac{B}{A}\)
\(P=\dfrac{4\sqrt{x}}{\sqrt{x}-2}:\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)
\(P=\dfrac{4\sqrt{x}}{\sqrt{x}-2}.\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\)
\(P=\dfrac{4\sqrt{x}}{\sqrt{x}+3}\)
\(P=4-\dfrac{12}{\sqrt{x}+3}\)
Để `P` nguyên thì \(\dfrac{12}{\sqrt{x}+3}\in Z\) hay \(\sqrt{x}+3\in U\left(12\right)=\left\{\pm1;\pm2;\pm4;\pm6;\pm12\right\}\)
Mà \(x\ge0\) \(\Rightarrow\sqrt{x}+3\ge3\) hay \(\sqrt{x}+3\in\left\{4;6;12\right\}\)
`@`\(\sqrt{x}+3=4\Rightarrow\sqrt{x}=1\Rightarrow x=1\left(tm\right)\)
`@`\(\sqrt{x}+3=6\Rightarrow\sqrt{x}=3\Rightarrow x=9\left(tm\right)\)
`@`\(\sqrt{x}+3=12\Rightarrow\sqrt{x}=9\Rightarrow x=81\left(tm\right)\)
Vậy \(x\in\left\{1;9;81\right\}\) thì `P` nguyên
