Question

tìm nghiệm của PT

\(\sqrt{x^2-9}+\sqrt{x-3}=0\)

Answer 1

ĐKXĐ: \(x\ge3\)

\(pt\Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}+\sqrt{x-3}=0\)

\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}+1\right)=0\)

\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(tm\right)\)(do \(\sqrt{x+3}+1\ge1>0\))

Answer 2

\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}+1\right)=0\)

hay x=3