Question

Giải phương trình: \(7+2\sqrt{x}-x=\left(2+\sqrt{x}\right)\sqrt{7-x}\)

Answer 1

ĐKXĐ: \(0\le x\le7\)

\(\Leftrightarrow7-x-\left(2+\sqrt{x}\right)\sqrt{7-x}+2\sqrt{x}=0\)

Đặt \(\sqrt{7-x}=t\ge0\)

\(t^2-\left(2+\sqrt{x}\right)t+2\sqrt{x}=0\)

\(\Delta=\left(2+\sqrt{x}\right)^2-8\sqrt{x}=\left(\sqrt{x}-2\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\frac{2+\sqrt{x}+\sqrt{x}-2}{2}=\sqrt{x}\\t=\frac{2+\sqrt{x}-\sqrt{x}+2}{2}=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{7-x}=\sqrt{x}\\\sqrt{7-x}=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=3\end{matrix}\right.\)