Question

Cho đẳng thức \(\frac{-3}{\sqrt{x}-4}=\sqrt{x}\). Giá trị x thỏa mãn đẳng thức

Answer 1

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne16\end{matrix}\right.\)

Ta có: \(\frac{-3}{\sqrt{x}-4}=\sqrt{x}\)

\(\Leftrightarrow-3=\sqrt{x}\cdot\left(\sqrt{x}-4\right)\)

\(\Leftrightarrow x-4\sqrt{x}+3=0\)

\(\Leftrightarrow x-\sqrt{x}-3\sqrt{x}+3=0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)-3\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1=0\\\sqrt{x}-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=9\left(nhận\right)\end{matrix}\right.\)

Vậy: S={1;9}