Question

Tìm x\(\in\)Z để

B=\(\left(\dfrac{\sqrt{x}+2}{2\sqrt{x}+x+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\) có giá trị nguyên

 

Answer 1

ĐK: x > 0

B = \(\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

⇔ B = \(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

⇔ B = \(\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\) = \(\dfrac{2}{x-1}\)

Để B ∈ Z thì x - 1 ∈ Ư(2) = {-2;-1;1;2}

⇔ \(\left\{{}\begin{matrix}x-1=-2\\x-1=-1\\x-1=1\\x-1=2\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}x=-3\\x=-2\\x=2\\x=3\end{matrix}\right.\)

Vậy....