điểu kiện x>=2
\(< =>\dfrac{\sqrt{x+3}^2-\sqrt{x-2}^2}{\sqrt{x+3}+\sqrt{x-2}}\left(1+\sqrt{\left(x+3\right)\left(x-2\right)}\right)=5\)
\(< =>\dfrac{5}{\sqrt{x+3}+\sqrt{x-2}}\left(1+\sqrt{\left(x+3\right)\left(x-2\right)}\right)=5\)
\(< =>\dfrac{1}{\sqrt{x+3}+\sqrt{x-2}}\left(1+\sqrt{\left(x+3\right)\left(x-2\right)}\right)=1\)
\(< =>1+\sqrt{\left(x+3\right)\left(x-2\right)}=\sqrt{x+3}+\sqrt{x-2}\)
đặt \(\sqrt{x+3}=a;\sqrt{x-2}=b\)
ta được \(1+ab=a+b< =>\left(a-1\right)\left(b-1\right)=0< =>\left[{}\begin{matrix}a=1\\b=1\end{matrix}\right.< =>\left[{}\begin{matrix}\sqrt{x+3}=1\\\sqrt{x-2}=1\end{matrix}\right.< =>\left[{}\begin{matrix}x=-2\left(ktm\right)\\x=3\left(tm\right)\end{matrix}\right.\)
