Question

Bai 1: A= \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\) B= \(\left(\dfrac{\sqrt{X}+1}{\sqrt{X}-1}-\dfrac{\sqrt{X}-1}{\sqrt{X}+1}\right)\) : \(\dfrac{\sqrt{X}}{\sqrt{X}-1}\) ( X> 0, X≠1)

A) Rut B

b) Tim x de gia tri cua A va B trai dau

(mink dag can rat gap)

 

Answer 1

a) Ta có: \(B=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}}{\sqrt{x}-1}\)

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

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

\(=\dfrac{4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

\(=\dfrac{4}{\sqrt{x}+1}\)

Answer 2

b. Để A và B trái dấu \(\Leftrightarrow AB< 0\)

\(\Leftrightarrow\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right)\left(\dfrac{4}{\sqrt{x}+1}\right)< 0\)

\(\Leftrightarrow\dfrac{4}{\sqrt{x}-1}< 0\Leftrightarrow\sqrt{x}-1< 0\)

\(\Rightarrow0< x< 1\)