ĐKXĐ: \(x\ne1;x\ne2\)
\(\frac{2x-1}{x-1}+\frac{3x-2}{x-2}=\frac{x^2+4x+5}{x^2-3x+2}+4\\ \Leftrightarrow\frac{2x^2-5x+2}{\left(x-1\right)\left(x-2\right)}+\frac{3x^2-5x+2}{\left(x-1\right)\left(x-2\right)}=\frac{x^2+4x+5}{\left(x-1\right)\left(x-2\right)}+\frac{4x^2-12x+8}{\left(x-1\right)\left(x-2\right)}\\ \Rightarrow2x^2-5x+2+3x^2-5x+2=x^2+4x+5+4x^2-12x+8\\ \Leftrightarrow2x^2+3x^2-x^2-4x^2-5x-5x-4x+12x=5-2-2\\ \Leftrightarrow-2x=1\\ \Leftrightarrow x=\frac{-1}{2}\left(tm\right)\)Vậy tập nghiệm của phương trình là: \(S=\left\{-\frac{1}{2}\right\}\)
ĐKXĐ: x∉{1;2}
Ta có: \(\frac{2x-1}{x-1}+\frac{3x-2}{x-2}=\frac{x^2+4x+5}{x^2-3x+2}+4\)
\(\Leftrightarrow\frac{\left(2x-1\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}+\frac{\left(3x-2\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}=\frac{x^2+4x+5}{\left(x-1\right)\left(x-2\right)}+\frac{4\left(x^2-3x+2\right)}{\left(x-1\right)\left(x-2\right)}\)
Suy ra: \(\left(2x-1\right)\left(x-2\right)+\left(3x-2\right)\left(x-1\right)=x^2+4x+5+4\left(x^2-3x+2\right)\)
\(\Leftrightarrow2x^2-4x-x+2+3x^2-3x-2x+2=x^2+4x+5+4x^2-12x+8\)
\(\Leftrightarrow5x^2-10x+4=5x^2-8x+13\)
\(\Leftrightarrow5x^2-10x+4-5x^2+8x-13=0\)
\(\Leftrightarrow-2x-9=0\)
\(\Leftrightarrow-2x=9\)
hay \(x=\frac{-9}{2}\)(tm)
Vậy: \(S=\left\{-\frac{9}{2}\right\}\)
