a. \(B=\dfrac{3x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\) (x ≥ 0; x ≠ 1; x ≠ 4)
\(=\dfrac{3x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3x-\left(x-1\right)+x-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\dfrac{3x-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\dfrac{3\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3\left(\sqrt{x}+1\right)}{\sqrt{x}-2}\)
b. \(P=A\cdot B=\dfrac{\sqrt{x}}{\sqrt{x}+1}\cdot\dfrac{3\left(\sqrt{x}+1\right)}{\sqrt{x}-2}=\dfrac{3\sqrt{x}}{\sqrt{x}-2}\)
Để \(\left|P\right|>P\) thì \(P< 0\Leftrightarrow\dfrac{3\sqrt{x}}{\sqrt{x}-2}< 0\)
\(\Rightarrow\left\{{}\begin{matrix}x>0\\\sqrt{x}-2< 0\left(\text{vì }3\sqrt{x}>0;\forall x>0\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>0\\\sqrt{x}< 2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>0\\x< 4\end{matrix}\right.\Leftrightarrow0< x< 4\)
Mà x nguyên nên \(x\in\left\{1;2;3\right\}\)
Kết hợp với ĐKXĐ của x, ta được: \(x\in\left\{2;3\right\}\)
\(\text{#}Toru\)
