a. \(\sqrt{x^2-3x+2}+\sqrt{x+3}=\sqrt{x^2+2x-3}+\sqrt{x-2}\)(ĐK; \(x\ge2\)
\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{\left(x+3\right)\left(x-1\right)}\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)+\sqrt{x+3}\left(1-\sqrt{x-1}\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x-2}-\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{x-2}-\sqrt{x+3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(TM\right)\\3=-2\left(VL\right)\end{matrix}\right.\)
Vậy x=2 là nghiệm của phương trình
