Question

Giải pt bằng cách đặt 1 ẩn t:

\(\sqrt{x}+\sqrt{9-x}=\sqrt{-x^2+9x+9}\)

Answer 1

ĐKXĐ: \(0\le x\le9\)

Bình phương 2 vế: \(9+2\sqrt{-x^2+9x}=-x^2+9x+9\)

Đặt \(\sqrt{-x^2+9x}=t\ge0\) pt trở thành:

\(t^2-2t=0\Rightarrow\left[{}\begin{matrix}t=0\\t=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{-x^2+9x}=0\\\sqrt{-x^2+9x}=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}-x^2+9x=0\\-x^2+9x-4=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=9\\x=\dfrac{9-\sqrt{65}}{2}\\x=\dfrac{9+\sqrt{65}}{2}\end{matrix}\right.\)