ĐKXĐ : \(x\ne2,x\ne4\)
Phương trình ban đầu tương đương :
\(\frac{x-1}{x-2}+\frac{x+3}{x-4}+\frac{2}{x^2-6x+8}=0\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x-4\right)+\left(x+3\right)\left(x-2\right)+2}{\left(x-2\right)\left(x-4\right)}=0\)
\(\Rightarrow x^2-5x+4+x^2+x-6+2=0\)
\(\Leftrightarrow2x^2-4x=0\)
\(\Leftrightarrow2x\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Rightarrow x=0\) ( Do x = 2 không thỏa mãn ĐKXĐ )
Vậy pt đã cho có tập nghiệm \(S=\left\{0\right\}\)
\(ĐKXĐ:x\ne2;x\ne4\)
\(\frac{x-1}{x-2}+\frac{x+3}{x-4}=\frac{2}{-x^2+6x-8}\)
\(\Rightarrow\frac{\left(x-1\right)\left(x-4\right)+\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=\frac{-2}{x^2-6x+8}\)
\(\Rightarrow\frac{\left(x^2-5x+4\right)+\left(x^2+x-6\right)}{x^2-6x+8}=\frac{-2}{x^2-6x+8}\)
\(\Rightarrow\frac{2x^2-4x-2}{x^2-6x+8}=\frac{-2}{x^2-6x+8}\)
\(\Rightarrow2x^2-4x-2=-2\)
\(\Rightarrow2x^2-4x=0\Rightarrow2x\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x=2\left(ktm\right)\end{cases}}\)
Vậy pt có 1 nghiệm duy nhất là 0
\(\frac{x-1}{x-2}+\frac{x+3}{x-4}=\frac{2}{-x^2+6x-8}\)
\(\Leftrightarrow\frac{x-1}{x-2}+\frac{x+3}{x-4}+\frac{2}{x^2-6x+8}=0\)
\(\Leftrightarrow\frac{x-1}{x-2}+\frac{x+3}{x-4}+\frac{2}{\left(x-2\right)\left(x-4\right)}=0\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}+\frac{\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}+\frac{2}{\left(x-2\right)\left(x-4\right)}=0\)
\(\Leftrightarrow\frac{x^2-5x+4}{\left(x-2\right)\left(x-4\right)}+\frac{x^2+x-6}{\left(x-2\right)\left(x-4\right)}+\frac{2}{\left(x-2\right)\left(x-4\right)}=0\)
\(\Leftrightarrow\frac{x^2-5x+4+x^2+x-6+2}{\left(x-2\right)\left(x-4\right)}=0\)
\(\Leftrightarrow2x^2+6x=0\)
\(\Leftrightarrow2x\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-3\end{cases}}}\)
Vậy x=0; x=-3
