Question

\(\sqrt{x+2+}\sqrt{5-x}+\sqrt{3x-x^{ }2+10}=4\)

Answer 1

ĐKXĐ: \(-2\le x\le5\)

Đặt \(\sqrt{x+2}+\sqrt{5-x}=t>0\)

\(\Rightarrow t^2=7+2\sqrt{3x-x^2+10}\Rightarrow\sqrt{3x-x^2+10}=\frac{t^2-7}{2}\) (1)

Pt trở thành:

\(t+\frac{t^2-7}{2}=4\Leftrightarrow t^2+2t-15=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-5\left(l\right)\end{matrix}\right.\)

Thế vào (1): \(\sqrt{3x-x^2+10}=\frac{t^2-7}{2}=1\)

\(\Leftrightarrow3x-x^2+10=1\Rightarrow x^2-3x-9=0\) (bấm máy)