Question

x+\(\sqrt{13-x^2}+x\sqrt{13-x^2}=1\)

Answer 1

ĐKXĐ: ...

Đặt \(x+\sqrt{13-x^2}=t\Rightarrow t^2=13+2x\sqrt{13-x^2}\)

\(\Rightarrow x\sqrt{13-x^2}=\frac{t^2-13}{2}\)

Pt trở thành:

\(t+\frac{t^2-13}{2}=1\Leftrightarrow t^2+2t-15=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+\sqrt{13-x^2}=3\\x+\sqrt{13-x^2}=-5\left(vn\right)\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{13-x^2}=3-x\left(x\le3\right)\)

\(\Leftrightarrow13-x^2=x^2-6x+9\)

\(\Leftrightarrow2x^2-6x-4=0\Rightarrow x=\frac{3-\sqrt{17}}{2}\)