điều kiện xác định : \(x\ge2\)
ta có : \(\dfrac{\sqrt{x+3}+\sqrt{x-2}}{\sqrt{x+3}-\sqrt{x-2}}=5-x^2\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x+3}+\sqrt{x-2}\right)^2}{\left(\sqrt{x+3}-\sqrt{x-2}\right)\left(\sqrt{x+3}+\sqrt{x-2}\right)}=5-x^2\) (vì \(\sqrt{x+3}+\sqrt{x-2}\ne0\forall x\) )
\(\Leftrightarrow\dfrac{x+3+x-2+2\sqrt{\left(x+3\right)\left(x-2\right)}}{x+3-x+2}=5-x^2\)
\(\Leftrightarrow\dfrac{2x+1+2\sqrt{x^2+x-6}}{5}=5-x^2\)
\(\Leftrightarrow2x+1+2\sqrt{x^2+x-6}=25-5x^2\)
\(\Leftrightarrow5x^2+2x-24+2\sqrt{x^2+x-6}=0\)
\(\Leftrightarrow\left(x-2\right)\left(5x+12\right)+2\sqrt{\left(x-2\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-2}\left(5x-12\right)+2\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\sqrt{x-2}\left(5x-12\right)+2\sqrt{x+3}=0\end{matrix}\right.\)
ta có : \(\sqrt{x-2}\left(5x-12\right)+2\sqrt{x+3}=0\) (1)
xảy ra khi \(5x-12< 0\Leftrightarrow2< x< \dfrac{12}{5}\)
\(\Rightarrow\left(1\right)\Leftrightarrow\left(12-5x\right)\sqrt{x-2}=2\sqrt{x+3}\)
bình phương 2 quế chuyển về 1 phía (cái pt này vô nghiệm thì phải)
nếu có nghiệm thì sử dụng sơ đồ hoocne nha
