\(\Leftrightarrow\frac{-x+1}{2}=\frac{x-2}{x-4}\)
\(\Leftrightarrow x^2+4x-3=2x-4\)
\(\Leftrightarrow x^2+2x+1=0\)
\(\Leftrightarrow\left(x+1\right)^2=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Bài này là bài lớp 8 mà.
Phương trình tương đương :
\(\frac{2\left(x+1\right)}{4}-x=\frac{x-2}{x-4}\)
<=> \(\frac{x+1}{2}-x=\frac{x-2}{x-4}\)
<=> \(\frac{x\left(x+1\right)}{2}-\frac{x-2}{x-4}=0\)
<=> \(\frac{x\left(x+1\right)\left(x-4\right)}{2\left(x-4\right)}-\frac{2\left(x-2\right)}{2\left(x-4\right)}=0\)
<=> \(\frac{x^3-4x^2+x^2-4x-2x+4}{2\left(x-4\right)}=0\)
<=> \(x^3-4x^2+x^2-4x-2x+4=0\)
<=> ....
