Question

A= \(\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x-1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{x}{x-1}\right)\)

a, rút gọn A

b, tìm \(x\in Z\) để \(A\in Z\)

c, tìm x để A đạt GTNN

Answer 1

a: 

Sửa đề: \(A=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}+2}{x\sqrt{x}+x-\sqrt{x}-1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{x}{x-1}\right)\)

\(A=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}+2}{\left(x-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}+1-x}{x-1}\)

\(=\dfrac{x-1-2\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(x-1\right)}\cdot\dfrac{x-1}{-x+\sqrt{x}+1}\)

\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)}\cdot\dfrac{1}{-x+\sqrt{x}+1}=\dfrac{-\sqrt{x}+3}{x-\sqrt{x}-1}\)

b: Để A là số nguyên thì \(\sqrt{x}\left(-\sqrt{x}+3\right)⋮x-\sqrt{x}-1\)

=>\(-x+3\sqrt{x}⋮x-\sqrt{x}-1\)

=>\(-x+\sqrt{x}+1+2\sqrt{x}-1⋮x-\sqrt{x}-1\)

=>\(x=0\)