Question

\(x+\sqrt{17-x^2}+x\sqrt{17-x^2}=9\)

Answer 1

ĐKXĐ: ...

Đặt \(x+\sqrt{17-x^2}=t\Rightarrow t^2=17+2x\sqrt{17-x^2}\)

\(\Rightarrow x\sqrt{17-x^2}=\frac{t^2-17}{2}\)

Pt trở thành:

\(t+\frac{t^2-17}{2}=9\Leftrightarrow t^2+2t-35=0\Rightarrow\left[{}\begin{matrix}t=5\\t=-7\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+\sqrt{17-x^2}=5\\x+\sqrt{17-x^2}=-7\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{17-x^2}=5-x\left(x\le5\right)\\\sqrt{17-x^2}=-7-x\left(vn\right)\end{matrix}\right.\)

\(\Leftrightarrow17-x^2=\left(5-x\right)^2\)

\(\Leftrightarrow2x^2-10x+8\Rightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)