\(Q=\dfrac{\sqrt{x}+6}{\sqrt{x}-2}\left(đk:x\ge0,x\ne4\right)=\dfrac{\sqrt{x}-2}{\sqrt{x}-2}+\dfrac{8}{\sqrt{x}-2}=1+\dfrac{8}{\sqrt{x}-2}\in Z\)
\(\Rightarrow\sqrt{x}-2\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\)
Do \(x\ge0,x\ne4\)
\(\Rightarrow x\in\left\{0;1;9;16;36;100\right\}\)
Đkxđ: x # 4
Q = 1 + 8/(sqrt(x) - 2)
Q nguyên --> sqrt(x) - 2 là ước của 8
Do sqrt(x) >=0 nên sqrt(x) - 2 >= -2
TH1: sqrt(x) - 2 = -2 <=> x = 0 (thỏa)
TH2: sqrt(x) - 2 = -1 <=> x = 1 (thỏa)
Th3: sqrt(x) - 2 = 1 <=> x = 9(thỏa)
TH4: sqrt(x) - 2 = 2<=> x = 16 (thỏa)
Th5: sqrt(x) - 2 = 4 <=> x = 36 (thỏa)
Th6: sqrt(x) - 2 = 8 <=> x = 100 (thỏa)
