\(a.K=\left(\dfrac{\sqrt{x}+2}{3\sqrt{x}}+\dfrac{2}{\sqrt{x}+1}-3\right):\dfrac{2-4\sqrt{x}}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1-x}{3\sqrt{x}}=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)+6\sqrt{x}-9\sqrt{x}\left(\sqrt{x}+1\right)}{3\sqrt{x}\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}+1}{2\left(1-2\sqrt{x}\right)}-\dfrac{3\sqrt{x}+1-x}{3\sqrt{x}}=\dfrac{2\left(1-2\sqrt{x}\right)\left(1+2\sqrt{x}\right)}{3\sqrt{x}}.\dfrac{1}{2\left(1-2\sqrt{x}\right)}-\dfrac{3\sqrt{x}+1-x}{3\sqrt{x}}=\dfrac{x-\sqrt{x}}{3\sqrt{x}}=\dfrac{\sqrt{x}-1}{3}\) \(b.x=\dfrac{1}{4}\left(KTMĐKXĐ\right)\) nên tại \(x=\dfrac{1}{4}\) giá trị của K không xác định .
\(c.K< 1\) ⇔ \(\dfrac{\sqrt{x}-1}{3}< 1\)
⇔ \(\sqrt{x}-1< 3\text{⇔}x< 16\)
Kết hợp với ĐKXĐ : \(0< x< 16\) ( x # \(\dfrac{1}{4}\) )
\(d.Để:\) K ∈ Z ⇔ \(\sqrt{x}-1\text{∈}\left\{1;-1;3;-3\right\}\)
+) \(\sqrt{x}-1=1\text{⇔ }x=4\left(TM\right)\)
+) \(\sqrt{x}-1=-1\text{⇔ }x=0\left(KTM\right)\)
+) \(\sqrt{x}-1=3\text{⇔ }x=16\left(TM\right)\)
+) \(\sqrt{x}-1=-3\text{⇔ }vô-nghiem\)
KL...............
