ĐKXĐ: \(\hept{\begin{cases}x^2+5x+3\ge0\\x^2+3x+2\ge0\end{cases}}\)
Đặt \(\hept{\begin{cases}\sqrt{x^2+5x+3}=a\\\sqrt{x^2+3x+2}=b\end{cases}\left(a;b\ge0\right)}\)
\(\Rightarrow a^2-b^2=x^2+5x+3-x^2-3x-2=2x+1\)
Pt trở thành
\(a-b=a^2-b^2\)
\(\Leftrightarrow a-b-\left(a-b\right)\left(a+b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(1-a-b\right)=0\)