\(\sqrt{x^2-5x+6}+\sqrt{x+1}=\sqrt{x-2}+\sqrt{x^2-2x-3}\)
ĐK: x^2-5x+6>=0<=> x<=2 hoặc x>=3
x^2-2x-3>=0<=> x<=-1 hoặc x>=3
<=>\(\sqrt{x^2-5x+6}+\sqrt{x+1}=\sqrt{x-2}+\sqrt{x^2-2x-3}\)
<=>\(\sqrt{\left(x-3\right)\left(x-2\right)}+\sqrt{x+1}=\sqrt{x-2}+\sqrt{\left(x-3\right)\left(x+1\right)}\)
<=> \(\sqrt{x-2}\left(\sqrt{x-3}-1\right)+\sqrt{x+1}\left(1-\sqrt{x-3}\right)=0\)
<=> \(\sqrt{x-2}\left(\sqrt{x-3}-1\right)-\sqrt{x+1}\left(\sqrt{x-3}-1\right)=0\)
<=> \(\left(\sqrt{x-3}-1\right)\left(\sqrt{x-2}-\sqrt{x+1}\right)=0\)
<=>\(\orbr{\orbr{\begin{cases}\sqrt{x-3}-1=0\\\sqrt{x-2}-\sqrt{x+1}=0\end{cases}}}\)
<=>\(\orbr{\begin{cases}\sqrt{x-3}=1\\\sqrt{x-2}=\sqrt{x+1}\end{cases}}\)
<=> \(\orbr{\begin{cases}x=4\left(nhan\right)\\0x=3\left(vôly\right)=>loai\end{cases}}\)
S={4}
