Question

Cho \(C=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}\)

Tìm x để \(\left|C\right|=\dfrac{1}{3}\)

Answer 1

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

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

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

Ta có: \(\left|C\right|=\dfrac{1}{3}\)

\(\Leftrightarrow C=\dfrac{-1}{3}\)

hay x=4

Answer 2

\(đk:x\ne1,x\ge0\)

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

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

\(\Rightarrow\sqrt{x}+1=3\Rightarrow\sqrt{x}=2\Rightarrow x=4\left(tm\right)\)