ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ne0\\x-2\ne0\\x-3\ne0\\x-4\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne1\\x\ne2\\x\ne3\\x\ne4\end{matrix}\right.\)
Ta có : \(\frac{x^2-2x+2}{x-1}+\frac{x^2-8x+20}{x-4}=\frac{x^2-4x+6}{x-2}+\frac{x^2-6x+12}{x-3}\)
=> \(\frac{x^2-2x+1+1}{x-1}+\frac{x^2-8x+16+4}{x-4}=\frac{x^2-4x+4+2}{x-2}+\frac{x^2-6x+9+3}{x-3}\)
=> \(\frac{\left(x-1\right)^2+1}{x-1}+\frac{\left(x-4\right)^2+4}{x-4}=\frac{\left(x-2\right)^2+2}{x-2}+\frac{\left(x-3\right)^2+3}{x-3}\)
=> \(x-1+\frac{1}{x-1}+x-4+\frac{4}{x-4}=x-2+\frac{2}{x-2}+x-3+\frac{3}{x-3}\)
=> \(\frac{1}{x-1}+\frac{4}{x-4}=\frac{2}{x-2}+\frac{3}{x-3}\)
=> \(\frac{x-4}{\left(x-1\right)\left(x-4\right)}+\frac{4\left(x-1\right)}{\left(x-4\right)\left(x-1\right)}=\frac{2\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+\frac{3\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}\)
=> \(\frac{\left(x-4+4\left(x-1\right)\right)}{\left(x-1\right)\left(x-4\right)}=\frac{2\left(x-3\right)+3\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
=> \(\frac{x-4+4x-4}{\left(x-1\right)\left(x-4\right)}=\frac{2x-6+3x-6}{\left(x-2\right)\left(x-3\right)}\)
=> \(\frac{5x-8}{x^2-5x+4}=\frac{5x-12}{x^2-5x+6}\)
=> \(5x^3-25x^2+30x-8x^2+40x-48=5x^3-25x^2+20x-12x^2+60x-48\)
=> \(5x^3-25x^2+30x-8x^2+40x-48-5x^3+25x^2-20x+12x^2-60x+48=0\)
=> \(4x^2-10x=0\)
=> \(2x\left(2x-5\right)=0\)
=> \(\left[{}\begin{matrix}2x=0\\2x-5=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=0\\x=\frac{5}{2}\end{matrix}\right.\) ( TM )
Vậy phương trình trên có tập nghiệm là \(S=\left\{0,\frac{5}{2}\right\}\)
