Bài 2:
ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\notin\left\{1;4\right\}\end{matrix}\right.\)
a: \(P=\left(\dfrac{6}{\sqrt{x}-1}-\dfrac{6}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
\(=\dfrac{6\sqrt{x}-6\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{6\sqrt{x}-6\sqrt{x}+6}{\left(\sqrt{x}-1\right)\cdot\sqrt{x}}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{x-1-\left(x-4\right)}\)
\(=\dfrac{6\left(\sqrt{x}-2\right)}{3\sqrt{x}}=\dfrac{2\sqrt{x}-4}{\sqrt{x}}\)
b: P=2
=>\(2\sqrt{x}-4=2\sqrt{x}\)
=>\(-4=0\left(vôlý\right)\)
Vậy: \(x\in\varnothing\)
c: Để P nguyên thì \(2\sqrt{x}-4⋮\sqrt{x}\)
=>\(-4⋮\sqrt{x}\)
=>\(\sqrt{x}\inƯ\left(-4\right)\)
mà \(\sqrt{x}>0\) với mọi x thỏa mãn ĐKXĐ
nên \(\sqrt{x}\in\left\{1;2;4\right\}\)
=>\(x\in\left\{1;4;16\right\}\)
Kết hợp ĐKXĐ, ta được: x=16
