Question

Cho biểu thức Q=\(\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x\sqrt{x}-x+\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}}{x+1}\right)\) (Với x\(\ge\)0, x\(\ne\)1)

a.Rút gọn Q

b. Chứng minh rằng Q\(\)>0

c.Tìm x để Q nguyên

Answer 1

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

\(ĐKXĐ:x\) ≥ \(0;x\) # \(1\)

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

\(Q=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(x+1\right)\left(\sqrt{x}-1\right)}.\dfrac{x+1}{x-\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{x-\sqrt{x}+1}\)

\(b.\) Ta thấy : \(x-\sqrt{x}+1=x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

Mà : \(\sqrt{x}+1>0\)

⇒ \(Q>0\)