\(\frac{x-3}{x-2}+\frac{x-2}{x-4}=-1\) (1)
ĐKXĐ \(x\ne2\) và \(x\ne4\)
\(\left(1\right)\Leftrightarrow\frac{x-2-1}{x-2}+\frac{x-4+2}{x-4}=-1\)
\(\Leftrightarrow1-\frac{1}{x-2}+1+\frac{2}{x-4}=-1\)
\(\Leftrightarrow2-\frac{1}{x-2}+\frac{2}{x-4}=-1\)
\(\Leftrightarrow\frac{1}{x-2}-\frac{2}{x-4}=3\)
\(\Leftrightarrow\frac{\left(x-4\right)-2\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=\frac{3\left(x-2\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}\)
\(\Rightarrow x-4-2x+4=3\left(x^2-6x+8\right)\)
\(\Leftrightarrow-x=3x^2-18x+24\)
\(\Leftrightarrow3x^2-18x+24+x=0\)
\(\Leftrightarrow3x^2-17x+24=0\)
\(\Leftrightarrow3x^2-9x-8x+24=0\)
\(\Leftrightarrow3x\left(x-3\right)-8\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(3x-8\right)=0\)
Th1 \(x-3=0\Leftrightarrow x=3\) (nhận)
Th2 \(3x-8=0\Leftrightarrow x=\frac{8}{3}\) (nhận)
Vậy Tập nghiệm của phương trình là \(S=\left\{3;\frac{8}{3}\right\}\)
