Ta có:
\(C=\left(\dfrac{1}{\sqrt{x}-1}-1\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\) (ĐK: \(x\ge0;x\ne1;x\ne4\))
\(C=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}-1}\right):\left[\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right]\)
\(C=\dfrac{1-\sqrt{x}+1}{\sqrt{x}-1}:\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(C=\dfrac{-\sqrt{x}+2}{\sqrt{x}-1}:\dfrac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(C=\dfrac{-\left(\sqrt{x}-2\right)}{\sqrt{x}-1}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{3}\)
\(C=\dfrac{-\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{3}\)
\(C=\dfrac{-\left(\sqrt{x}-2\right)^2}{3}\)
\(C=-3\) khi:
\(\dfrac{-\left(\sqrt{x}-2\right)^2}{3}=-3\)
\(\Leftrightarrow-\left(\sqrt{x}-2\right)^2=-9\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-2=3\\\sqrt{x}-2=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=5\\\sqrt{x}=-1\left(L\right)\end{matrix}\right.\)
\(\Leftrightarrow x=25\left(tm\right)\)
