ĐKXĐ : \(x^2-5x+6\ne0\)
=> \(\left(x-2\right)\left(x-3\right)\ne0\)
=> \(\left[{}\begin{matrix}x\ne2\\x\ne3\end{matrix}\right.\)
- Ta có : \(\left(x-2\right)\left(x-3\right)-3\sqrt{x^2-5x+6}=2\)
=> \(\left(x-2\right)\left(x-3\right)-3\sqrt{x^2-2x-3x+6}=2\)
=> \(\left(x-2\right)\left(x-3\right)-3\sqrt{x\left(x-2\right)-3\left(x-2\right)}=2\)
=> \(\left(x-2\right)\left(x-3\right)-3\sqrt{\left(x-2\right)\left(x-3\right)}=2\)
Đặt \(a=\sqrt{\left(x-2\right)\left(x-3\right)}\left(a\ge0\right)\) ta được phương trình :
\(a^2-3a=2\)
=> \(a^2-3a-2=0\)
=> \(a^2-\frac{2.x.3}{2}+\frac{9}{4}-\frac{17}{4}=0\)
=> \(\left(a-\frac{3}{2}\right)^2-\left(\sqrt{\frac{17}{4}}\right)^2=0\)
=> \(\left(a-\frac{3}{2}-\sqrt{\frac{17}{4}}\right)\left(a-\frac{3}{2}+\sqrt{\frac{17}{4}}\right)=0\)
=> \(\left[{}\begin{matrix}a-\frac{3}{2}-\sqrt{\frac{17}{4}}=0\\a-\frac{3}{2}+\sqrt{\frac{17}{4}}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}a=\frac{3}{2}+\sqrt{\frac{17}{4}}=\frac{3+\sqrt{17}}{2}\left(TM\right)\\a=\frac{3}{2}-\sqrt{\frac{17}{4}}=\frac{3-\sqrt{17}}{2}\left(KTM\right)\end{matrix}\right.\)
=> \(a=\frac{3+\sqrt{17}}{2}\)
- Thay \(a=\sqrt{\left(x-2\right)\left(x-3\right)}\) vào phương trình trên ta được :
\(\sqrt{\left(x-2\right)\left(x-3\right)}=\frac{3+\sqrt{17}}{2}\)
=> \(\left(x-2\right)\left(x-3\right)=\frac{13+3\sqrt{17}}{2}\)
=> \(x^2-2x-3x+6=\frac{13+3\sqrt{17}}{2}\)
=> \(\frac{2x^2}{2}-\frac{4x}{2}-\frac{6x}{2}+\frac{12}{2}=\frac{13+3\sqrt{17}}{2}\)
=> \(2x^2-10x+12=13+3\sqrt{17}\)
=> \(2x^2-10x-1-3\sqrt{17}=0\)
=> \(\left(x\sqrt{2}\right)^2-2.x.\sqrt{2}.\frac{5\sqrt{2}}{2}+\frac{25}{2}-\frac{27+6\sqrt{17}}{2}=0\)
=> \(\left(x\sqrt{2}-\frac{5\sqrt{2}}{2}\right)^2=\frac{27+6\sqrt{17}}{2}\)
=> \(\left[{}\begin{matrix}x\sqrt{2}-\frac{5\sqrt{2}}{2}=\sqrt{\frac{27+6\sqrt{17}}{2}}\\x\sqrt{2}-\frac{5\sqrt{2}}{2}=-\sqrt{\frac{27+6\sqrt{17}}{2}}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x\sqrt{2}=\frac{5\sqrt{2}}{2}+\sqrt{\frac{27+6\sqrt{17}}{2}}\\x\sqrt{2}=\frac{5\sqrt{2}}{2}-\sqrt{\frac{27+6\sqrt{17}}{2}}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{\frac{5\sqrt{2}}{2}+\sqrt{\frac{27+6\sqrt{17}}{2}}}{\sqrt{2}}\\x=\frac{\frac{5\sqrt{2}}{2}-\sqrt{\frac{27+6\sqrt{17}}{2}}}{\sqrt{2}}\end{matrix}\right.\)
Vậy phương trình có nghiệm là \(\left[{}\begin{matrix}x=\frac{\frac{5\sqrt{2}}{2}+\sqrt{\frac{27+6\sqrt{17}}{2}}}{\sqrt{2}}\\x=\frac{\frac{5\sqrt{2}}{2}-\sqrt{\frac{27+6\sqrt{17}}{2}}}{\sqrt{2}}\end{matrix}\right.\) .
