Question

Giúp tớ nheeeee

Answer 1

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

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

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

b: Để B>1/3 thì B-1/3>0

=>\(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{3}>0\)

=>\(\dfrac{3-\sqrt{x}-1}{3\left(\sqrt{x}+1\right)}>0\)

=>\(2-\sqrt{x}>0\)

=>\(\sqrt{x}< 2\)

=>0<=x<4

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0< =x< 4\\x\ne1\end{matrix}\right.\)