Ta có: \(\sqrt{x^2-3x+2}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{x^2+2x-3}\)
<=>\(\sqrt{\left(x^2-x\right)-\left(2x-2\right)}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{\left(x^2-x\right)+\left(3x-3\right)}\)<=> \(\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x+3}-\sqrt{x-2}-\sqrt{\left(x+3\right)\left(x-2\right)}\) = 0
<=> \(\sqrt{\left(x-2\right)}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)\) = 0
<=> \(\left(\sqrt{x-2}+\sqrt{x+3}\right)\left(\sqrt{x-1}-1\right)=0\)
<=> \(\left[{}\begin{matrix}\sqrt{x-2}+\sqrt{x+3}=0\\\sqrt{x-1}-1=0\end{matrix}\right.\)
=> \(\sqrt{x-1}=1\) ( Do \(\sqrt{x-2}+\sqrt{x+3}\) vô nghiệm)
=> \(x-1=1\) => x = 2
Vậy ...............................
P/s : Tui mới lớp 8 ak! Nên có sai đừng trách nha!
