Ta có: \(\left\{{}\begin{matrix}x^2-4x+3=\left(x-1\right)\left(x-3\right)\\x^2-6x+8=\left(x-2\right)\left(x-4\right)\end{matrix}\right.\) hay: \(pt\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x-2\right)\left(x-4\right)=15\) \(\Rightarrow\left(x-1\right)\left(x-4\right)\left(x-2\right)\left(x-3\right)=15\) \(\Rightarrow\left(x^2-4x-x+4\right)\left(x^2-3x-2x+6\right)=15\) \(\Rightarrow\left(x^2-5x+4\right)\left(x^2-5x+6\right)=15\) \(\Rightarrow\left(x^2-5x+5-1\right)\left(x^2-5x+5+1\right)=15\) \(\Rightarrow\left(x^2-5x+5\right)^2-16=0\Leftrightarrow\left(x^2-5x+5+4\right)\left(x^2-5x+5-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x^2-5x+9=0\\x^2-5x+1=0\end{matrix}\right.\) Xét: \(x^2-5x+9=0\Leftrightarrow x^2-5x+\dfrac{25}{4}+\dfrac{11}{4}=0\Leftrightarrow\left(x-\dfrac{5}{2}\right)^2+\dfrac{11}{4}=0\)(vô nghiệm) Xét: \(x^2-5x+1=0\Leftrightarrow x^2-5x+\dfrac{25}{4}-\dfrac{21}{4}=0\Leftrightarrow\left(x-\dfrac{5}{2}\right)^2-\dfrac{21}{4}=0\Leftrightarrow\left(x-\dfrac{5}{2}+\sqrt{\dfrac{21}{4}}\right)\left(x-\dfrac{5}{2}-\sqrt{\dfrac{21}{4}}\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}-\sqrt{\dfrac{21}{4}}\\x=\dfrac{5}{2}+\sqrt{\dfrac{21}{4}}\end{matrix}\right.\)
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