Question

[hello hello]

1. cho biểu thức A=\(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}:\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}\)

a, rút gọn

b, tìm giá trị nguyễn của x để A giá trị nguyên

Answer 1

[l҉o҉n҉g҉ d҉z҉]

ĐKXĐ : x > 0 , x khác 1

\(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right)\div\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}\right)\)

\(=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\div\left[\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\)

\(=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\div\left[\dfrac{x+\sqrt{x}-x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\)

\(=2\div\dfrac{1}{\sqrt{x}-1}=2\sqrt{x}-2\)

b) Dễ thấy ∀ x ≥ 0 thì \(2\sqrt{x}-2\) nguyên

Kết hợp với ĐKXĐ => Với \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)thì A đạt giá trị nguyên