Question

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

Answer 1

ĐK: \(x\ge3\)

\(\sqrt{x^2-x-6}=\sqrt{x-3}\Leftrightarrow\left(\sqrt{x^2-x-6}\right)^2=\left(\sqrt{x-3}\right)^2\Leftrightarrow x^2-x-6=x-3\Leftrightarrow x^2-2x-3=0\Leftrightarrow x^2-3x+x-3=0\Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)

Vậy S={3}