Question

 

1,\(\sqrt{2+\sqrt{\text{x}-5}}=\sqrt{13-x}\)

Answer 1

ĐKXĐ: \(5\le x\le13\)

Đặt \(\sqrt{x-5}=t\ge0\Rightarrow x=t^2+5\)

Phương trình trở thành:

\(\sqrt{2+t}=\sqrt{13-\left(t^2+5\right)}\)

\(\Leftrightarrow\sqrt{t+2}=\sqrt{8-t^2}\)

\(\Leftrightarrow t+2=8-t^2\)

\(\Leftrightarrow t^2+t-6=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-3\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x-5}=2\)

\(\Rightarrow x=9\)