Question

Rút gọn biểu thức:

B=\(\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{x+1}{x-1}\) với x\(\ge\)0 và x\(\ne\)+-1.Tìm x để B<1

Answer 1

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

\(=\dfrac{2}{x-1}.\dfrac{x-1}{x+1}=\dfrac{2}{x+1}\)

Để \(B< 1\Rightarrow\dfrac{2}{x+1}< 1\Rightarrow1-\dfrac{2}{x+1}>0\Rightarrow\dfrac{x-1}{x+1}>0\)

mà \(x+1>0\left(x\ge0\right)\Rightarrow x-1>0\Rightarrow x>1\)

Answer 2

Answer 3

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

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

\(=\dfrac{2}{x+1}\)

Answer 4

b) Để B<1 thì B-1<0

\(\Leftrightarrow\dfrac{2}{x+1}-1< 0\)

\(\Leftrightarrow\dfrac{2-x-1}{x+1}< 0\)

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

\(\Leftrightarrow\dfrac{x-1}{x+1}>0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1>0\\x+1< 0\end{matrix}\right.\Leftrightarrow x>1\)