ĐK: \(x\ge3\)
\(\sqrt{x^2-5x+6}+\sqrt{x+1}=\sqrt{x-2}+\sqrt{x^2-2x-3}\)
<=>\(\sqrt{\left(x-2\right)\left(x-3\right)}+\sqrt{x+1}=\sqrt{x-2}+\sqrt{\left(x+1\right)\left(x-3\right)}\)
<=>\(\sqrt{\left(x-2\right)\left(x-3\right)}+\sqrt{x+1}-\sqrt{x-2}-\sqrt{\left(x+1\right)\left(x-3\right)}=0\)
<=> \(\sqrt{x-2}\left(\sqrt{x-3}-1\right)-\sqrt{x+1}\left(\sqrt{x-3}-1\right)=0\)
<=> \(\left(\sqrt{x-3}-1\right)\left(\sqrt{x-2}-\sqrt{x+1}\right)=0\)
Đến đây dễ rồi bn tự làm tiếp nhé
\(\sqrt{x^2-5x+6}+\sqrt{x+1}=\sqrt{x-2}+\sqrt{x^2-2x-3}\) (ĐK: \(x\ge3\) )
\(\Leftrightarrow\sqrt{x^2-2x-3x+6}+\sqrt{x+1}=\sqrt{x-2}+\sqrt{x^2+x-3x-3}\)
\(\Leftrightarrow\sqrt{x\left(x-2\right)-3\left(x-2\right)}+\sqrt{x+1}=\sqrt{x-2}+\sqrt{x\left(x+1\right)-3\left(x+1\right)}\)
\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x-3\right)}+\sqrt{x+1}-\sqrt{x-2}-\sqrt{\left(x+1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\left(\sqrt{\left(x-2\right)\left(x-3\right)}-\sqrt{x-2}\right)-\left(\sqrt{\left(x+1\right)\left(x-3\right)}-\sqrt{x+1}\right)=0\)
\(\Leftrightarrow\left[\sqrt{x-2}.\sqrt{x-3}-\sqrt{x-2}\right]-\left[\sqrt{x+1}.\sqrt{x-3}-\sqrt{x+1}\right]=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-3}-1\right)-\sqrt{x+1}\left(\sqrt{x-3}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-\sqrt{x+1}\right)\left(\sqrt{x-3}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-2}-\sqrt{x+1}=0\\\sqrt{x-3}-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{x-2}=\sqrt{x+1}\\\sqrt{x-3}=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x-2=x+1\\x-3=1\end{cases}}}\)
\(\Leftrightarrow\orbr{\begin{cases}0x=3\Rightarrow x\in\varnothing\\x=4\Rightarrow TM\end{cases}}\)
Vậy x = 4
P/s: sai sót xin bỏ qua.
