Cho biết: \(\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}=2\). Hỏi x = ?
\(x=8\). \(x=\dfrac{1}{4}\). \(x=2\sqrt{2}\). \(x=16\). Hướng dẫn giải:Đkxđ: \(x\ne4\)
\(\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}=2\)
\(\Leftrightarrow\dfrac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=2\)
\(\Leftrightarrow\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=2\)
\(\Leftrightarrow\dfrac{\sqrt{x}\cdot\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=2\)
\(\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}=2\)
\(\Leftrightarrow\sqrt{x}=2\left(\sqrt{x}-2\right)\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=16\).
