Question

giải phương trình \(6\sqrt{x+2}+3\sqrt{3-x}=3x+1+4\sqrt{-x^2+x+6}\)

Answer 1

ĐKXĐ: \(-2\le x\le3\)

Đặt \(2\sqrt{x+2}+\sqrt{3-x}=a>0\)

\(\Rightarrow a^2=3x+11+4\sqrt{-x^2+x+6}\)

Phương trình trở thành:

\(3a=a^2-10\Leftrightarrow a^2-3a-10=0\Rightarrow\left[{}\begin{matrix}a=5\\a=-2< 0\left(l\right)\end{matrix}\right.\)

\(\Rightarrow2\sqrt{x+2}+\sqrt{3-x}=5\)

\(\Leftrightarrow3x+11+4\sqrt{-x^2+x+6}=25\)

\(\Leftrightarrow4\sqrt{-x^2+x+6}=14-3x\)

\(\Leftrightarrow25\left(x^2-4x+4\right)=0\)

\(\Rightarrow x=2\)