Lời giải:
\(x> 0; x\neq 1\)
a) Đặt \(\sqrt{x}=a\). Ta có:
\(P=\left(\frac{a^3-1}{a^2-a}-\frac{a^3+1}{a^2+a}\right):\left(\frac{2(a^2-2a+1)}{a^2-1}\right)\)
\(=\left(\frac{(a-1)(a^2+a+1)}{a(a-1)}-\frac{(a+1)(a^2-a+1)}{a(a+1)}\right):\frac{2(a-1)^2}{(a-1)(a+1)}\)
\(=\left(\frac{a^2+a+1}{a}-\frac{a^2-a+1}{a}\right):\frac{2(a-1)}{a+1}\)
\(=2.\frac{a+1}{2(a-1)}=\frac{a+1}{a-1}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b) \(P=\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{x}-1+2}{\sqrt{x}-1}=1+\frac{2}{\sqrt{x}-1}\)
Để P là số nguyên thì \(\Leftrightarrow \frac{2}{\sqrt{x}-1}\in\mathbb{Z}\Leftrightarrow \sqrt{x}-1\in \left\{\pm 1;\pm 2\right\}\)
\(\Leftrightarrow \sqrt{x}\in\left\{0;2;3\right\}\)
\(\Leftrightarrow x\in \left\{0;4;9\right\}\). Mà $x\neq 0$ nên \(x\in \left\{4;9\right\}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne0\\x\ne1\end{matrix}\right.\)
\(P=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left[\dfrac{2\left(x-2\sqrt{x}+1\right)}{x-1}\right]\)
\(=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]:\left[\dfrac{2\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\right]\)
\(=\left(\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\right):2\left(\sqrt{x}-1\right)\)
\(=2\times\dfrac{1}{2\left(\sqrt{x}-1\right)}\)
\(=\dfrac{1}{\sqrt{x}-1}\)
b )
Để P đạt giá trị nguyên thì :
\(1⋮\sqrt{x}-1\) hay \(\sqrt{x}-1\) là ước của 1 .
\(Ư\left(1\right)=\left(-1;1\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}-1=1\\\sqrt{x}-1=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\left(N\right)\\x=0\left(L\right)\end{matrix}\right.\)
Vậy \(x=4\)
